Suppose you have chain complexes $A,B,C,D$. And say you have maps $f:A \to B, g:B \to C, h:C \to D$, but only $g$ is a chain map. However, $h\circ g\circ f$ is a chain map even though $f$ and $h$ weren't chain maps.
Now suppose that $g$ induces the $0$ map on Homology. Then does it follow in some way that $h\circ g\circ f$ induces the $0$ map on Homology?
Let $A\cong B\cong D$ be $1$ dimensional concentrated in a single degree, say $0$. Let $C$ be of the form $0\to C_2\to C_1\to 0$ where $C_2$ and $C_1$ are both $1$ dimensional and the boundary map is an isomorphism. Let $g,h,f$ just carry the generators in degree $1$ to each other. Then clearly the composition $h\circ g\circ f$ is a chain map and the induced map on homology is the identity. Moreover $g$ is also clearly a chain map. So this is a counterexample to the conjecture.