Let $E$ be the set of all event length strings over $\sigma^*$ which is regular as we can easily get a DFA for this. Now, $L = R \cap E$ and since intersection being closed for regular set, $L$ must be a regular set.
Since $R$ is regular we can have a DFA for $R$. Now, mark all the final states of the DFA and identify the ones where a string is reachable with even length. At this point we can have states where both the strings of even/odd length can be reached and in such cases we have to replicate the state (reverse of what we do in DFA minimization). After this we get a DFA for $L$ and so $L$ is always regular.
PS: Sorry, the proof is more of an intuition and not mathematical.