Recall the acyclic models theorem: given two functors $F, G$ from a "category $\mathcal{C}$ with models $M$" to the category of chain complexes of modules over a ring $R$, a natural transformation $H_0(F) \to H_0(G)$ induces a natural transformation $F \to G$ (unique up to natural chain homotopy) if $F$ is free on the models and $G$ is acyclic. (This has many useful applications, such as showing that things like the Alexander-Whitney maps exist and are unique up to homotopy.)

The proof is a bit of homological algebra, but I suspect that there may be a fancier homotopical way to think about it. Namely, I want to say something of the form that the category of functors from $\mathcal{C}$ to chain complexes has a model structure on it, the free functors are cofibrant in some sense, and the map $G \to H_0(G)$ should be an acyclic fibration. Thus the lifting of $F \to H_0(F) \to H_0(G)$ to $F \to G$ would just be a lifting argument in a general model category (as would be the existence of a natural chain homotopy). However, I'm not sure what the model structure should be. The projective model structure doesn't seem to be the right one because $G \to H_0(G)$ is not a weak equivalence of chain complexes (it only is for the models $M$). So I guess the model structure relevant here would be a hybrid of the projective model structure, where instead of weak equivalences and fibrations levelwise, one would just want them to hold on $M$: that is, a morphism $A \to B$ is a weak equivalence (resp. fibration) if and only if $A(m) \to B(m)$ is one for each $m \in M$. It seems that the lifting property needed is essentially the proof of the acyclic models theorem itself.

I strongly suspect this can be done. Am I right? Can this be pushed further?