This is a rather lazy version of an answer, but I will also indicate how I think a full answer can be produced.
Let $u(x,z)$ be the solution of $-u''+qu=z u$ with $u(0,z)=0$, $u'(0,z)=1$. Then
$$
B_L = \left\{ F(z)=\int f(x)u(x,z)\, dx: f\in L^2(0,L) \right\}
$$
is a Hilbert space of entire functions with a reproducing kernel (a de Branges space). You are asking if there is an $F\in B_{1/2}$ satisfying $F(\lambda_{n^2}(1;q))=a_n$, with $\{\lambda_n(L;q)\}$ denoting the spectrum of the Dirichlet operator $-d^2/dx^2+q(x)$ on $[0,L]$ (equivalenty, the zeros of $u(L,z)$).
This is an interpolation problem. The space $B_L$ as a set does not depend on the potential $q$. This is proved (for example) in my paper on Schrodinger operators and de Branges spaces; see reference 24 here.
This already gives us some information: If $\{\lambda_{n^2}(1;q)\}$ is a subset of $\{\lambda_k(1/2;q_1)\}$ for some $q_1$, then interpolation in $B_{1/2}$ is possible.
Moreover, since the possibility of interpolation typically depends on the density of the sequence, we should be very safe in your situation, when $\lambda_{n^2}(1)\simeq n^4\pi^2$, while $\lambda_n(1/2)\simeq 4n^2\pi^2$. In fact, this suggests that it would also work on $\{ \lambda_{2n}(1;q)\}$ or at least close to this situation.
To actually prove something, I think it would be a good idea to relate
$$
B_L = \left\{ F(z) =\int g(x) \frac{\sin\sqrt{z}x}{\sqrt{z}}\, dx : g\in L^2(0,L) \right\} ;
$$
to the Paley-Wiener space $PW_L$. For example, if there is no negative spectrum (which we can always achieve by adding a constant to $q$) and we change variables $G(z)=zF(z^2)$, then we are now dealing with the odd part of the Paley Wiener space $PW_L$.
Interpolation in Paley-Wiener spaces has been studied extensively and a literature search should produce the information we need here. Your setting has the additional complication that we don't know the interpolation points $\lambda_{n^2}$ explicitly, but that should not be an issue because we do know their asymptotic behavior $\lambda_{n^2}\simeq n^4\pi^2$.