Protocol for Randomized Oblivious Transfer?
No, as written, your protocol doesn't work -- the problem is that Bob is supposed to be allowed to choose $b$, your protocol selects a random one for him.

However, it is close -- here is a modification that I believe does work:

First, suppose Alice has her values $(x_0, x_1)$, and Bob has his bit $b$.

They run their Random functionality R, and so Alice gets the values $(z_0, z_1)$, and Bob gets the values $(c, z_c)$.

Now, Bob sends the Alice the bit $e$ defined as $e = b \oplus c$; Alice sends back a pair of values $(y_0, y_1)$ defined as $(y_0, y_1) = (x_0 \oplus z_e, x_1 \oplus z_{1 \oplus e})$.

Bob then computes $z_c \oplus y_b = x_b$.

I claim that Alice gets no information about the value of $b$, and that Bob gets no information about the value of $(x_{1 \oplus b})$, and thus this is a valid Oblivious Transfer function.