Let $\mathcal{C}$ be a model category with a forgetful functor towards simplicial sets, and such that fibrations and trivial cofibrations are those whose underlying map on simplicial sets is a fibration, respectively trivial fibration. Is there a criterion to decide whether formation of simplicial homotopy groups commutes with filtered colimits?
I'm not aware of any such criterion (and I'm doubtful a very general one exists, if any), so this is not meant to be an answer (but for some reason I'm unable to post it as a comment right now).
A few remarks. The question is slightly ill posed. What do you mean by "commutation of simplicial homotopy groups with filtered colimits"? You probably mean formation of simplicial homotopy groups turns, in degree zero, filtered colimits in $\mathcal{C}$ into filtered colimits of pointed sets, and in positive degree it turns filtered colimits in $\mathcal{C}$ into filtered colimits of groups.
It seems to me a necessary condition on $\mathcal{C}$ is that it is proper (see GoerssJardine, Ch. II, $\S$8). Indeed, if formation of higher simplicial homotopy groups commutes with filtered colimits in the above sense, then in particular pushouts along cofibrations preserve weak equivalences.
In fact, a sharper necessary condition is that weak equivalences in $\mathcal{C}$ are closed under filtered colimits. This should be seen as a strong "finiteness condition" on $\mathcal{C}$: it is typically met if generating cofibrations and generating trivial cofibrations are small relative to the whole category $\mathcal{C}$ (which is stronger than requiring $\mathcal{C}$ to be "finitely generated").
You may then ask whether it is the case that in a model category satisfying your stated assumptions, and in which, in addition, filtered colimits preserve weak equivalences, formation of higher simplicial homotopy groups commutes with filtered colimits in the above sense.
I don't have a counterexample off the top of my head, but I'm led to believe this is not the case in general.

2$\begingroup$ Thanks! I found a counterexample thanks to these remarks. I'll type it up as time permits, and post it here. $\endgroup$ – user95222 Oct 6 '17 at 9:37