Yes, application of $U$ to Alice's half of $|\beta\rangle=(|00\rangle+|11\rangle)/\sqrt{2}$ and appliaction of $U^T$ to Bob's half are equivalent. In fact the identity
$$
(A \otimes I)|\beta\rangle = (I \otimes A^T) |\beta\rangle
$$
is true for any matrix (not necessarily unitary or real). Let
$$
A = \begin{pmatrix}
a_{00} & a_{01} \\
a_{10} & a_{11}
\end{pmatrix}
$$
and compare
$$
(A \otimes I)|\beta\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}
a_{00} & 0 & a_{01} & 0 \\
0 & a_{00} & 0 & a_{01} \\
a_{10} & 0 & a_{11} & 0 \\
0 & a_{10} & 0 & a_{11} \\
\end{pmatrix} \cdot \begin{pmatrix}
1 \\ 0 \\ 0 \\ 1
\end{pmatrix} =
\frac{1}{\sqrt{2}}\begin{pmatrix}
a_{00} \\ a_{01} \\ a_{10} \\ a_{11}
\end{pmatrix}
$$
with
$$
(I \otimes A^T) |\beta\rangle = \begin{pmatrix}
a_{00} & a_{10} & 0 & 0 \\
a_{01} & a_{11} & 0 & 0 \\
0 & 0 & a_{00} & a_{10} \\
0 & 0 & a_{01} & a_{11}
\end{pmatrix} \cdot \begin{pmatrix}
1 \\ 0 \\ 0 \\ 1
\end{pmatrix} =
\frac{1}{\sqrt{2}}\begin{pmatrix}
a_{00} \\ a_{01} \\ a_{10} \\ a_{11}
\end{pmatrix}.
$$
A more elegant way to prove the identity uses tensor networks, see example 9 and equation $(46)$ on page 12 in this paper.