Let $f,g : A\rightrightarrows B$ be two maps between chain complexes $A,B$ with terms in an abelian category. Suppose $f,g$ are homotopic. Must there exist a homotopy equivalence $\alpha : C\rightarrow A$ such that $f\circ \alpha = g\circ \alpha$ on the nose? (also, the same question with $\beta : B\rightarrow C$ such that $\beta\circ f = \beta\circ g$)
No. For instance, in the category of vector spaces over a field $k$, let $A$ be the chain complex $0\to k\to 0$ (with $k$ in degree $0$) and let $B$ be the chain complex $0\to k\to k\to 0$ (with the $k$'s in degrees $1$ and $0$). Let $f$ be the inclusion map $A\to B$ and let $g$ be the zero map $A\to B$. Then $f$ and $g$ are homotopic, since $B$ is contractible. But if $\alpha:C\to A$ is any homotopy equivalence, then $\alpha$ must be surjective (in order to be surjective on homology), and so $f\alpha$ and $g\alpha$ cannot be equal.
For homotopy equivalences $B\to C$, you can use the same example but in the opposite category. (Or, the quotient map and the zero map $B\to B/A$ give a counterexample for the category of vector spaces, since any homotopy equivalence $B/A\to C$ must be injective.)