In the discussions about quantum correlations, particularly beyond entanglement (discord, dissonance e.t.c), one can often meet two definitions of mutual information of a quantum system $\rho^{AB}$: $$ I(\rho^{AB}) = S(\rho^A) + S(\rho^B) - S(\rho^{AB}) $$ and $$ J(\rho^{AB}) = S(\rho^A)-S_{\{\Pi^B_j\}}(\rho^{A|B}), $$ where $S$ is the Von-Neumann entropy, $\rho^A$ and $\rho^B$ are the reduced states of the individual subsystems of $\rho^{AB}$ and the second term in $J$ is the quantum analogue of the conditional entropy $$ S_{\{\Pi^B_j\}}(\rho^{A|B}) = \sum_j p_j S(\rho^{A|\Pi^B_j}). $$ In the expression for the conditional entropy $\rho^{A|\Pi^B_j} = \text{Tr}_B[\rho^{AB} (\mathbb{I}^A \otimes \Pi^B_j )]/p_j $ are the states of the subsystem $A$ after getting a particular projector $\Pi^B_j$ in $B$, which happens with a probability $p_j = \text{Tr}[\rho^{AB} (\mathbb{I}^A \otimes \Pi^B_j ) ]$. While $I$ characterizes the total correlations between $A$ and $B$ the second expression involves a measurement process, in which non-classical features of $\rho^{AB}$ are lost, and therefore $J$ characterizes classical correlations in $\rho^{AB}$.
While measuring $J$ is relatively straightforward, (for 2 qubits one can just measure 4 probabilities $p(\Pi^A_i \Pi^B_j), \, i,j = 1,2$ and calculate the mutual information of the resulting probability distribution) I can't think of an easy way of estimating $I$. So my question is: is it possible to measure $I$ without performing a full tomography of $\rho^{AB}$?