No counterexamples exist. Indeed, given a polynomial $P$ consider the recursive sequence $b_{n+1}=f(b_n)$ where I take $f(x)=x+P(x)$, say. Then $P(b_{n+1}) = P(b_n + P(b_n)) \equiv P(b_n) \equiv 0 \bmod P(b_n)$ since $x-y \mid P(x)-P(y)$ in general. Setting $a_n=P(b_n) \in R_P$ this says that $a_{n+1}$ is divisible by $a_n$.

One can replace $f$ by more general polynomials. An important property of $f$ is that it permutes the roots of $P$.

Every $P$ is a counterexample. Indeed, given a polynomial $P$ consider the recursive sequence $b_{n+1}=f(b_n)$ where I take $f(x)=x+P(x)$, say. Then $P(b_{n+1}) = P(b_n + P(b_n)) \equiv P(b_n) \equiv 0 \bmod P(b_n)$ since $x-y \mid P(x)-P(y)$ in general. Setting $a_n=P(b_n) \in R_P$ this says that $a_{n+1}$ is divisible by $a_n$.

One can replace $f$ by more general polynomials. An important property of $f$ is that it permutes the roots of $P$.