This problem is from Aluffi's Algebra: Chapter 0. Let $R$ be an integral domain that is not a field, and let $F$ be a free $R$-module of finite rank. If $\varphi:F\to F$ is an endomorphism, what property of $R$ immediately guarantees that $\ker \varphi^{n}=\ker\varphi^{n+1}$, for $n\gg 0$?
If we look at the proof for vector spaces over a field, it seems to me that it relies on one important fact. This fact is that finite dimensional vector spaces are characterized by its dimension. The thing is, that this is true for free modules over commutative rings (IBN property). Is this the property that guarantees stabilizing kernels? I'm not sure since I am not using that $R$ is an integral domain.
I suspect the property that is being asked for here is that of being Artinian, since if $R$ is Artinian, it satisfies the descending chain condition, so all its finitely generated modules do too, so the descending chain condition implies that $ker(\varphi)\supset ker(\varphi^2)\supset ker(\varphi^3)\supset..$ must stabilise, giving the result.
But the claim is true for all (commutative) integral domains $R$, tensor your map up to the quotient field to $Q(R)$ to get an endomorphism of a finite dimensional vector space, so the kernel will stabilise, and then the kernel of your original map $\varphi^n$ is just the intersection of $ker((\varphi \otimes Q(R))^n)$ with $F$ inside $F\otimes Q(R)$.