Security of hash functions over input extended with a random value
I think so cause $M, R$ can be assumed to be as $X$, so in essence $H(X)$ is being hashed and the birthday attack possibility still exists.


Yes, that's correct.

Note that there is no such function as $H(M, R)$, you'd need some way of combining the values; lets assume concatenation though (so $H(M | R)$).