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).

Take the category of right modules for the matrix ring $B=\begin{pmatrix}A&A\\0&A\end{pmatrix}$, so that an object can be thought of as a pair of $A$-modules $(M,N)$ and a map $(M,N)\to(M',N')$ consists of $A$-module maps $M\to M'$, $N\to N'$ and $M\to N'$, and take the (non-full) subcategory with the same objects, but only those maps where $M\to N'$ has finite rank. This is an additive category $\mathcal{A}$.

For $k>0$, let $X(k)$ be the complex
$$\dots\to 0\to (A,0)\stackrel{\varepsilon}{\to}(A,0)\stackrel{\varepsilon}{\to}\dots\stackrel{\varepsilon}{\to}(A,0)\stackrel{\varepsilon}{\to}(A,0)\to0\to\dots$$
with $(A,0)$ in degrees $-k,\dots,0$, and let $Y(k)$ be the complex
$$\dots\to 0\to (0,A)\stackrel{\varepsilon}{\to}(0,A)\stackrel{\varepsilon}{\to}\dots\stackrel{\varepsilon}{\to}(0,A)\stackrel{\varepsilon}{\to}0\to0\to\dots$$
with $(0,A)$ in degrees $-k,\dots,-1$.

Consider the chain map $f(k):X(k)\to Y(k)$ that is induced by $\varepsilon:A\to A$ in degree $-k$ 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 $-k,\dots,0$.