No, it is not always attained at a positive semidefinite matrix. The simplest example I have been able to find to demonstrate this is as follows:

\begin{align*}
    U = \{ (1,0), (0,1), \tfrac{1}{\sqrt{2}}(1,1), \tfrac{1}{\sqrt{2}}(1,-1) \}, \quad \mathbf{x} = (1,2)/\sqrt{5}.
\end{align*}

For this optimization problem, the optimal value is $6/5$ (easily found by LP solvers), and here is an example of a Hermitian matrix attaining this value:
\begin{align*}
H = \begin{bmatrix}
0 & 1/2 \\ 1/2 & 1
\end{bmatrix}.
\end{align*}

However, there is no PSD matrix attaining the same value $6/5$. I don't see a "nice" way to see this, but if you add the constraint that $H$ is PSD, then it is now a semidefinite program (and thus solvable), and now has an optimal value of $(1+\sqrt{2})^2/5 \approx 1.165\ldots \leq 6/5 = 1.2$. This could be proved rigorously by constructing the dual of the SDP if desired.