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 $-k,\dots,0$.

Probably there's a much less artificial example, and even more probably there's a much simpler one, but ...

Let $A=k[\varepsilon]/(\varepsilon^2)$, where $k$ is a field of 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 whose 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 with only those maps $\alpha$ where $\alpha_2=0$.

For $k\geq0$, let $X(k)$ be the chain 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 (over $\mathcal{A}$) $f(k):X(k)\to Y(k)$ 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 given by $\begin{pmatrix}0&0\\1&0\end{pmatrix}$ in degrees $k,\dots,0$), but there's no choice of contracting homotopy with degree zero component in $\mathcal{A}_0$.

Now let $\mathcal{A}^{\mathbb{N}}$ be the category of $\mathbb{N}$-graded objects of $\mathcal{A}$, and consider the chain map $$(f(k))_{k\in\mathbb{N}}:(X(k))_{k\in\mathbb{N}}\to (Y(k))_{k\in\mathbb{N}}.$$

As a map of chain complexes over $\mathcal{A}^{\mathbb{N}}$ this is null-homotopic.

However, if you regard it a map of chain complexes over the (non-full) subcategory of $\mathcal{A}^{\mathbb{N}}$ with the same objects but where maps are required to have grade $d$ components in $\mathcal{A}_0$ for all but finitely many $d$, then it isn't null-homotopic but (using contracting homotopies for finitely many $f(k)$) is homotopic to a map that is zero in any given finite set of degrees.

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 $-k,\dots,0$.

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 $-k,\dots,0$.