Is $L(G) \subseteq L(R)$ decidable?
Yes, your thought is correct. The only missing part, as you pointed out, is whether or how we can create a CFL algorithmically for the intersection of a CFL with a RL.

If you take a close look at any proof for the fact that the intersection of a CFL with a RL is a CFL, you will find that the proof is constructive or can be made to be constructive easily, giving that there is an algorithm to construct the CFL for a given PDA. Here "constructive" means the same as "algorithmic". It is just by convention or history, we tend to use the word "constructive" instead of "algorithmic" to describe a proof.

You can check your textbook, or this and this.

Exercise. Is the following problem decidable?


  Given a context-free grammar $G$ and a regular expression $R$, is $L(G) \supseteq L(R)$?