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 $6/5$. To see this, write \begin{align*} H = \begin{bmatrix} a & b \\ \overline{b} & d \end{bmatrix}. \end{align*} The constraints of the problem imply $0 \leq a,d \leq 1$ and $0 \leq a + 2\mathrm{Re}(b) + d \leq 1$, and to get an objective function value of $6/5$ we would need $\mathbf{x}^*H\mathbf{x} = (a + 4\mathrm{Re}(b) + 4d)/5 = 6/5$, so $a = 6 - 4\mathrm{Re}(b) - 4d$. Plugging this back into one of the earlier constraints, we see $0 \leq 6 - 2\mathrm{Re}(b) - 3d \leq 1$. Since $d \leq 1$ and positive semidefiniteness implies $\mathrm{Re}(b) \leq 1$, this actually forces $d = \mathrm{Re}(b) = 1$, so $a = 6-4-4=-2$, which contradicts positive semidefiniteness.