I like the question very much.

First, let me mention briefly that the question has a flaw in the quantifier order, since you have first fixed the theory $\Gamma$ and then ask for a cardinal $\kappa$ such that if all subtheories $\Delta\subset\Gamma$ of size at most $\kappa$ are consistent, then $\Gamma$ is consistent. This is trivially affirmative, since we may simply let $\kappa=|\Gamma|$, in which case $\Delta=\Gamma$ is one of the allowed subtheories.

The actual question here is the following (and note that I replace your $\leq\kappa$ with $\lt\kappa$, since this is how one usually frames it with weakly and strongly compact cardinals):

Question. Is there are a cardinal $\kappa$ such that if $\Gamma$ is any $L_{\omega_1,\omega}$ theory in any signature, and every $\kappa$-small subtheory is consistent, then $\Gamma$ is consistent?

Let us call this property the $\kappa$-compactness property for $L_{\omega_1,\omega}$. Just to be clear, $L_{\omega_1,\omega}$ is the infinitary language in which one is allowed to form countable conjunctions and disjunctions, but still only finitely many quantifiers at a time. Meanwhile, $L_{\kappa,\lambda}$ allows conjunctions and disjunctions of size less than $\kappa$ and blocks of quantifiers of size less than $\lambda$. One instinctively thinks of the following large cardinals:

• A cardinal $\kappa$ is weakly compact if and only if it is uncountable $L_{\kappa,\kappa}$ has the $\kappa$-compactness property for theories in a language of size at most $\kappa$.

• A cardinal $\kappa$ is strongly compact if and only if $L_{\kappa,\kappa}$ has the $\kappa$-compactness property for any theory without any size restriction.

One crucial difference between $L_{\omega_1,\omega}$ and $L_{\kappa,\kappa}$ or even $L_{\omega_1,\omega_1}$ is that in $L_{\omega_1,\omega_1}$, one can express the assertion that a relation is well-founded, since you can say that it has no infinite descending sequence. This does not seem possible to express in $L_{\omega_1,\omega}$, because one can quantify only finitely many variables at a time.

Theorem. If $\kappa$ is strongly compact, then $L_{\omega_1,\omega}$ has the $\kappa$-compactness property.

Proof. This is immediate, since any $L_{\omega_1,\omega}$ theory is also a $L_{\kappa,\kappa}$ theory. QED

Theorem. If $L_{\omega_1,\omega}$ has the $\kappa$-compactness property, then there is a measurable cardinal.

Proof. Suppose that $L_{\omega_1,\omega}$ has the $\kappa$-compactness property. Let $\Gamma$ be the theory including the following assertions:

• the full $L_{\omega_1,\omega}$ diagram of the structure $\langle \kappa,\in,\hat A\rangle_{A\subset \kappa}$, in the language with a predicate $\hat A$ for each $A\subset\kappa$ and constants $\hat\alpha$ for each $\alpha\in\kappa$.
• the assertions $c\neq \hat\alpha$ for each $\alpha\in\kappa$.

Note that any $\kappa$-small subtheory of $\Gamma$ is consistent, since we may interpret $c$ inside $\kappa$ if only fewer than $\kappa$ many $\alpha$ are excluded. So by the $\kappa$-compactness property, $\Gamma$ has a model $\langle M,\hat\in,\hat A^M\rangle$. Let $U$ be the set of $A$ for which $M\models c\in\hat A$. This $U$ is an ultrafilter and it is countably complete, since the assertions $(\forall x. \bigwedge_n x\in A_n)\to x\in A$, whenever $A=\cap A_n$, are part of $\Gamma$. It is nonprincipal since $c\neq \hat\alpha$ for any $\alpha$. So there is a countably complete nonprincipal ultrafilter, and hence there is a measurable cardinal, since the degree of completeness of such an ultrafilter is always measurable. QED

In particular, the hypothesis is strictly stronger than a weakly compact cardinal.

I'll give some more thought to the exact strength, which I think might be the strongly compact cardinal.