Probably there's a much less artificial example, but ...
Let $A=k[\varepsilon]/(\varepsilon^2)$, where $k$ is a field that I'll take to have characteristic two (purely so I don't need to bother about signs).
Let $\mathcal{A}$ be the category whose objects are pairs $(M,N)$ of $A$-modules and maps $\alpha:(M,N)\to(M',N')$ are matrices $\begin{pmatrix}\alpha_1&0\\\alpha_2&\alpha_3\end{pmatrix}$ where $\alpha_1:M\to M'$, $\alpha_2:M\to N'$, $\alpha_3:N\to N'$ are $A$-module homomorphisms.
Let $\mathcal{A}_0$ be the (non-full) subcategory with the same objects, but only those maps $\alpha$ where $\alpha_2=0$.
For $k\geq0$, let $X(k)$ be the complex $$\dots\to 0\to (A,0)\to (A,0)\to\dots\to(A,0)\to(A,0)\to0\to\dots$$ with $(A,0)$ in degrees $-k-1,\dots,0$, where all the non-trivial differentials are $\begin{pmatrix}\varepsilon&0\\ 0&0\end{pmatrix}$, and let $Y(k)$ be the complex $$\dots\to 0\to (0,A)\to (0,A)\to\dots\to(0,A)\to0\to0\to\dots$$ with $(0,A)$ in degrees $-k-1,\dots,-1$, where all the non-trivial differentials are $\begin{pmatrix}0&0\\ 0&\varepsilon\end{pmatrix}$.
Consider the chain map $f(k):X(k)\to Y(k)$ given by $\begin{pmatrix}0&0\\ \varepsilon&0\end{pmatrix}$ in degree $-k-1$ and zero in other degrees. This is null-homotopic (for example, take the homotopy induced by $1:A\to A$ in degrees $-k+1,\dots,0$), but any chain homotopy involves a non-zero map $(A,0)\to(0,A)$ in degree zero.
Now take $\bigoplus_{k>0}X(k)$ and $\bigoplus_{k>0}Y(k)$, and the chain map $f$between them induced by the $f(k)$. This is a map of complexes over $\mathcal{A}$, since each component has rank one. It is null-homotopic as a chain map over the category of $B$-modules, by taking the sum of homotopies for each individual $f(k)$, but any chain homotopy involves a map of infinite rank in degree zero, so it is not null-homotopic over $\mathcal{A}$.
However, taking the sum of homotopies between $f(1),\dots,f(k)$ and the zero map for any $k>0$ gives a homotopy between $f$ and a map that is zero in degrees $-k,\dots,0$.