Answer to edited question:

It's still not true for qubit systems. Consider these two unit vectors, both of which are entangled: $$ |\phi\rangle = \frac{1}{\sqrt{2}} | 00\rangle + \frac{1}{\sqrt{2}} | 11\rangle,\\ |\psi\rangle = \frac{1}{\sqrt{2}} | 01\rangle + \frac{i}{\sqrt{2}} | 10\rangle. $$ Let $\rho = |\psi\rangle\langle \psi |$, which of course is not on the boundary of the set of separable states; it's as far away from that set as any state can be. However, we have $$ \text{Tr}\Bigl( |\phi\rangle \langle \phi |^{T_2} \rho\Bigr) = 0. $$ To verify this equation, it may help to view the relevant operators as matrices: $$ |\phi\rangle \langle \phi |^{T_2} = \frac{1}{2}\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix} $$ and $$ \rho = |\psi\rangle \langle \psi | = \frac{1}{2}\begin{pmatrix}0&0&0&0\\0&1&i&0\\0&-i&1&0\\0&0&0&0\end{pmatrix}. $$

Original answer:

If I understand it correctly, the statement suggested in the question is false.

For a counter-example, let $\rho$ be any entangled PPT state whose partial transpose has an entangled state $|\phi\rangle$ in its kernel. We then obtain $$ \text{Tr}\Bigl( |\phi\rangle \langle \phi |^{T_2} \rho\Bigr) = \text{Tr}\Bigl( |\phi\rangle \langle \phi | \rho^{T_2}\Bigr) = 0, $$ but because $\rho$ is entangled it cannot be on the boundary of the set of separable states.

Incidentally, no entanglement witness of the form suggested in the question can ever detect entanglement in a PPT state.

To construct an entangled PPT state $\rho$ whose partial transpose has an entangled state in its kernel, one may take any unextendable product basis $$ |\phi_1\rangle |\psi_1\rangle,\ldots,|\phi_m\rangle |\psi_m\rangle \in \mathcal{A}\otimes\mathcal{B}, $$ and then let $$ \rho = \frac{1}{m} \sum_{k = 1}^m |\phi_k\rangle \langle \phi_k| \otimes |\psi_k\rangle \langle \psi_k|. $$ The partial transpose looks like this: $$ \rho^{T_2} = \frac{1}{m} \sum_{k = 1}^m |\phi_k\rangle \langle \phi_k| \otimes |\overline{\psi_k}\rangle \langle \overline{\psi_k}|. $$ This operator has rank $m<\text{dim}(\mathcal{A}\otimes\mathcal{B})$ and every nonzero vector in its kernel is entangled.

Answer to edited question:

It's still not true for qubit systems. Consider these two unit vectors, both of which are entangled: $$ |\phi\rangle = \frac{1}{\sqrt{2}} | 00\rangle + \frac{1}{\sqrt{2}} | 11\rangle,\\ |\psi\rangle = \frac{1}{\sqrt{2}} | 01\rangle + \frac{i}{\sqrt{2}} | 10\rangle. $$ Let $\rho = |\psi\rangle\langle \psi |$, which of course is not on the boundary of the set of separable states; it's as far away from that set as any state can be. However, we have $$ \text{Tr}\Bigl( |\phi\rangle \langle \phi |^{T_2} \rho\Bigr) = 0. $$ To verify this equation, it may help to view the relevant operators as matrices: $$ |\phi\rangle \langle \phi |^{T_2} = \frac{1}{2}\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix} $$ and $$ \rho = |\psi\rangle \langle \psi | = \frac{1}{2}\begin{pmatrix}0&0&0&0\\0&1&-i&0\\0&i&1&0\\0&0&0&0\end{pmatrix}. $$

Original answer:

If I understand it correctly, the statement suggested in the question is false.

For a counter-example, let $\rho$ be any entangled PPT state whose partial transpose has an entangled state $|\phi\rangle$ in its kernel. We then obtain $$ \text{Tr}\Bigl( |\phi\rangle \langle \phi |^{T_2} \rho\Bigr) = \text{Tr}\Bigl( |\phi\rangle \langle \phi | \rho^{T_2}\Bigr) = 0, $$ but because $\rho$ is entangled it cannot be on the boundary of the set of separable states.

Incidentally, no entanglement witness of the form suggested in the question can ever detect entanglement in a PPT state.

To construct an entangled PPT state $\rho$ whose partial transpose has an entangled state in its kernel, one may take any unextendable product basis $$ |\phi_1\rangle |\psi_1\rangle,\ldots,|\phi_m\rangle |\psi_m\rangle \in \mathcal{A}\otimes\mathcal{B}, $$ and then let $$ \rho = \frac{1}{m} \sum_{k = 1}^m |\phi_k\rangle \langle \phi_k| \otimes |\psi_k\rangle \langle \psi_k|. $$ The partial transpose looks like this: $$ \rho^{T_2} = \frac{1}{m} \sum_{k = 1}^m |\phi_k\rangle \langle \phi_k| \otimes |\overline{\psi_k}\rangle \langle \overline{\psi_k}|. $$ This operator has rank $m<\text{dim}(\mathcal{A}\otimes\mathcal{B})$ and every nonzero vector in its kernel is entangled.

If I understand it correctly, the statement suggested in the question is false.

For a counter-example, let $\rho$ be any entangled PPT state whose partial transpose has an entangled state $|\phi\rangle$ in its kernel. We then obtain $$ \text{Tr}\Bigl( |\phi\rangle \langle \phi |^{T_2} \rho\Bigr) = \text{Tr}\Bigl( |\phi\rangle \langle \phi | \rho^{T_2}\Bigr) = 0, $$ but because $\rho$ is entangled it cannot be on the boundary of the set of separable states.

Incidentally, no entanglement witness of the form suggested in the question can ever detect entanglement in a PPT state.

To construct an entangled PPT state $\rho$ whose partial transpose has an entangled state in its kernel, one may take any unextendable product basis $$ |\phi_1\rangle |\psi_1\rangle,\ldots,|\phi_m\rangle |\psi_m\rangle \in \mathcal{A}\otimes\mathcal{B}, $$ and then let $$ \rho = \frac{1}{m} \sum_{k = 1}^m |\phi_k\rangle \langle \phi_k| \otimes |\psi_k\rangle \langle \psi_k|. $$ The partial transpose looks like this: $$ \rho^{T_2} = \frac{1}{m} \sum_{k = 1}^m |\phi_k\rangle \langle \phi_k| \otimes |\overline{\psi_k}\rangle \langle \overline{\psi_k}|. $$ This operator has rank $m<\text{dim}(\mathcal{A}\otimes\mathcal{B})$ and every nonzero vector in its kernel is entangled.

Answer to edited question:

It's still not true for qubit systems. Consider these two unit vectors, both of which are entangled: $$ |\phi\rangle = \frac{1}{\sqrt{2}} | 00\rangle + \frac{1}{\sqrt{2}} | 11\rangle,\\ |\psi\rangle = \frac{1}{\sqrt{2}} | 01\rangle + \frac{i}{\sqrt{2}} | 10\rangle. $$ Let $\rho = |\psi\rangle\langle \psi |$, which of course is not on the boundary of the set of separable states; it's as far away from that set as any state can be. However, we have $$ \text{Tr}\Bigl( |\phi\rangle \langle \phi |^{T_2} \rho\Bigr) = 0. $$ To verify this equation, it may help to view the relevant operators as matrices: $$ |\phi\rangle \langle \phi |^{T_2} = \frac{1}{2}\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix} $$ and $$ \rho = |\psi\rangle \langle \psi | = \frac{1}{2}\begin{pmatrix}0&0&0&0\\0&1&i&0\\0&-i&1&0\\0&0&0&0\end{pmatrix}. $$

Original answer:

If I understand it correctly, the statement suggested in the question is false.

For a counter-example, let $\rho$ be any entangled PPT state whose partial transpose has an entangled state $|\phi\rangle$ in its kernel. We then obtain $$ \text{Tr}\Bigl( |\phi\rangle \langle \phi |^{T_2} \rho\Bigr) = \text{Tr}\Bigl( |\phi\rangle \langle \phi | \rho^{T_2}\Bigr) = 0, $$ but because $\rho$ is entangled it cannot be on the boundary of the set of separable states.

Incidentally, no entanglement witness of the form suggested in the question can ever detect entanglement in a PPT state.

To construct an entangled PPT state $\rho$ whose partial transpose has an entangled state in its kernel, one may take any unextendable product basis $$ |\phi_1\rangle |\psi_1\rangle,\ldots,|\phi_m\rangle |\psi_m\rangle \in \mathcal{A}\otimes\mathcal{B}, $$ and then let $$ \rho = \frac{1}{m} \sum_{k = 1}^m |\phi_k\rangle \langle \phi_k| \otimes |\psi_k\rangle \langle \psi_k|. $$ The partial transpose looks like this: $$ \rho^{T_2} = \frac{1}{m} \sum_{k = 1}^m |\phi_k\rangle \langle \phi_k| \otimes |\overline{\psi_k}\rangle \langle \overline{\psi_k}|. $$ This operator has rank $m<\text{dim}(\mathcal{A}\otimes\mathcal{B})$ and every nonzero vector in its kernel is entangled.