If $L$ is context-free and $R$ is regular, then $L / R$ is context-free?
Here is a hint.

You need your machine to initially accept part of a word from $L$, consuming the tape as it goes. Then, without consuming anything, you need to find some word from $R$ that will push the machine into a final state. The chosen word from $R$ plays the role of the input word for the second half of the computation.

Clearly, non-determinism will have a role, as will the product between the two machines. The trick in formalising this is adjusting the product to deal with the fact that the input comes from $R$ not from the input.