Is the following true: If two chain complexes of free abelian groups have isomorphic homology modules then they are chain homotopy equivalent.

Yes, this is true. Suppose $C_*$ is such a chain complex of free abelian groups. For each $n$, choose a splitting of the boundary map $C_n \to B_{n1}$, so that $C_n \cong Z_n \oplus B_{n1}$. (You can do this because $B_{n1}$, as a subgroup of a free group, is free.) For all $n$, you then have a subchaincomplex $\cdots \to 0 \to B_n \to Z_n \to 0 \to \cdots$ concentrated in degrees $n$ and $n+1$, and $C_*$ is the direct sum of these chain complexes. Given two such chain complexes $C_*$ and $D_*$, you get a direct sum decomposition of each, and so it suffices to show that any two complexes $\cdots \to 0 \to R_i \to F_i \to 0 \to \cdots$, concentrated in degrees $n$ and $n+1$, which are resolutions of the same module $M$ are chain homotopy equivalent; but this is some variant of the fundamental theorem of homological algebra. This is special to abelian groups and is false for modules over a general ring. 


Yes, this is true, and it does not matter whether the complexes are bounded from any side (nor of course does it matter whether the homology is finitely generated). This is so because:



The natural functor $K^b(\mathbb Z\mathrm{free})\to D^b(\mathbb Z)$ from the homotopy category of bounded complexes of finitely generated free abelian groups to the derived category of bounded complexes of finitely generated abelian groups is an equivalence. This means that a map of bounded complexes of finitely generated free abelian groups which induces an isomorphism in homology is an homotopy equivalence. This and the fact that one can always lift a morphism $f:H_\bullet(X)\to H_\bullet(Y)$ between the homologies of two complexes of free abelian groups to a morphism $\tilde f:X\to Y$ of complexes which induces $f$ give an affirmative answer to your question. 

