For what it's worth, here's as much of a proof as I can work out

$\displaystyle P(A\cap B|R)=\frac{P(A\cap B\cap R)}{P(R)}$ by one of the definitions of a conditional probability. And since by the same definition

$\displaystyle P(A|R)*P(B|R)=\frac{P(A\cap R)*P(B\cap R)}{P(R)*P(R)}$ The only way that

$\displaystyle P(A\cap B|R)=P(A|R)*P(B|R)$ under the assumption that A and B are independent is if

$\displaystyle P(A\cap B\cap R) \stackrel{?}{=}\frac{P(A\cap R)*P(B\cap R)}{P(R)}$ under the same assumption. And you can use the same identity in the opposite direction to reduce the right side to

$\displaystyle P(A\cap B\cap R) \stackrel{?}{=} P(A|R)*P(B\cap R)$ which I'm not sure about