Yes, but the answer is a bit unsatisfying. This answer is a summary of the very nice paper Computable Fields and Galois Theory, Russel Miller, *Notices of the AMS*, 2008.

First of all, if one could not even compute with the elements of the ring $R$ at all, it would be unclear what it would mean for factorization to be computable. The usual solution is to talk about "computable rings", meaning a countable (or finite) ring $R$ where the elements are indexed by integers and the operations of addition and multiplication are given by computable functions.

The following is an example of a computable field in which factorization is not decidable: Let $p_n$ be the $n$-th prime and let $T_n$ be the $n$-th Turing machine and let $K = \mathbb{Q}(\sqrt{p_n} \ : \ T_n \ \mbox{halts})$. Given any element $\theta$ of $K$, there is a finite expression which witnesses that $\theta$ is in $K$: namely, the algebraic expression for $\theta$ in terms of finitely many $p_n$'s and a transcript of the ruunning of the corresponding Turing machines. This can be made into a proof that $K$ is computable, in particular (in response to Matt F.'s comment) there is no issue with testing equality. However, $x^2 - p_n$ is reducible iff $T_n$ halts, so we cannot test reducibility.

The fields one normally meets in life do not have this issue. Reducibility is computable over $\mathbb{Q}$, $\mathbb{R}$ and over finite fields. If reducibility is computable over $k$, then it is computable over $k(t)$ and over $k[t]/f(t)$ for $f$ irreducible. That already covers the most obvious fields you know.

Searching for "computable fields" turns up a fair bit of recent research.