Does every (co)homology functor (in particular, stable homotopy) factor through chain complexes?

No (I mean, not in a triangulated way), otherwise any generalized homology theory of a mod 2 Moore space would be 2-torsion, but this is not true for mod 2 stable homotopy groups (it's well known that you get a cyclic group of order 4). For positive results under extra hypotheses see:

Heller, A., 1966. Extraordinary Homology and Chain Complexes, in: Proceedings of the Conference on Categorical Algebra, La Jolla. pp. 355–365.

Neeman, A., 1992. Stable homotopy as a triangulated functor. Invent Math 109, 17–40. doi:10.1007/BF01232016

As written by Fernando, the answer is no if you suppose that you have exactness fo pairs at the chain level. In fact for a functor $$L_*:CW^{pairs}\rightarrow Ch(Ab)$$ if you have a short exact sequence of chain complexes $$0\rightarrow L_*(A,\emptyset)\rightarrow L_*(X,\emptyset)\rightarrow L_*(X,A)\rightarrow 0$$ and if the homology $\mathcal{L_*}=H_*(L_*(-))$ is a generalized homology theory then we have:

Theorem (Burdick, Conner, Floyd 1968).To each finite CW-pair (X,A) we have a natural isomorphism: $$\mathcal{L}_n(X,A)\cong \bigoplus_{p+q=n} H_p(X,A;\mathcal{L}_q(pt)).$$

Ref: "Chain Theories and their Derived Homologies", Proc . Amer. Math. Soc. l9 (l968) Proceedings of the AMS (1968)