Let me preface this by saying that I'm confident that the answer to the question in the title is "No," but I'm looking for an example to see why.
Let $\mathcal{A}$ be an abelian category. Let $\operatorname{Kom}(\mathcal{A})$ be the category of chain complexes over $\mathcal{A}$, $K(\mathcal{A})$ be the homotopy category of chain complexes, and $D(\mathcal{A})$ the derived category. Let $X$ be a chain complex over $\mathcal{A}$. Let $$\operatorname{End}_{\operatorname{Kom}}(X), \; \operatorname{End}_{K(\mathcal{A})}(X), \; \operatorname{End}_{D(\mathcal{A})}(X)$$ be the respective endomorphism rings. By definition, $\operatorname{End}_{K(\mathcal{A})}$ is a quotient of $\operatorname{End}_{\operatorname{Kom}}$, and $\operatorname{End}_{D(\mathcal{A})}$ is a localization of $\operatorname{End}_{K(\mathcal{A})}$. I have computed these rings explicitly for some complexes of abelian groups, but in every example I've done, the quasi-isomorphisms within $\operatorname{End}_{K(\mathcal{A})}$ are all units, so the localization has no effect.
I'd like to find an example of a chain complex $X$ of abelian groups so that $\operatorname{End}_{K(\mathcal{A})} \neq \operatorname{End}_{D(\mathcal{A})}$. Can anyone provide such an example?
I'm also interested in examples of complexes $X,Y$ where $\operatorname{Hom}_{K(\mathcal{A})} \neq \operatorname{Hom}_{D(\mathcal{A})}$.
EDIT: Someone marked this as a duplicate of Quasi-isomorphism and homotopical equivalence, but that question is different from mine and the answers there do not answer my question. In the example chain complex $X$ provided by Mike Miller, the non-nullhomotopic elements of $\operatorname{End}_{K(\mathcal{A})}(X)$ are all invertible, so localizing by quasi-isomorphisms has no effect. That is, $\operatorname{End}_{K(\mathcal{A})} = \operatorname{End}_{D(\mathcal{A})}$.