The question is exactly that of the title. Suppose $\varphi\in V$ is an $\mathcal{L}_{\infty\omega}$-sentence, and $W$ is an inner model of $V$ such that $\varphi\in W$. Is the statement

$\varphi$ has a model

absolute between $V$ and $W$? It is clearly upwards absolute, by induction on rank, so I'm really asking about the downwards direction.

The main obstacle that I see right now is the lack of a Lowenheim-Skolem theorem, which prevents us from using Shoenfield absoluteness. As a specific example, let $A$ be $\omega_1^W$ as a linear order, and suppose $V$ is a forcing extension of $W$ in which $\omega_1$ is collapsed. In $V$, there is a sentence $\varphi\in\mathcal{L}_{\omega_1\omega}\subset\mathcal{L}_{\infty\omega}$ whose only countable model up to isomoprhism is $A$: the Scott sentence of $A$. Now there is a formula $\psi\in \mathcal{L}_{\infty\omega}\cap W$ such that, in $V$, $\psi$ and $\varphi$ are equivalent; this is a roundabout way of saying "$\varphi\in W$," which is not exactly true. But then $\psi$ has no countable models in $W$, since such a model would have to be isomorphic to $A$ in $V$.

I assume this is well-known, but I haven't been able to find an answer myself.