No, just take $A$ and $B$ to be intervals, then it's clearly not true that $\mu(A \cap T^n B) - \mu(A) \mu(B)$ is small on average since $T^n$ is uniformly distributed among rotations of the unit circle (for all $n>0$) and $\mu(A \cap T^n B) - \mu(A) \mu(B)$ is large for some positive fraction of rotations.

No,

$$ \frac{1}{n} \sum_{k=1}^{n}| \mu (A) \cap T^{-k}(B) - \mu(A)\mu(B)|$$ is a sum of i.i.d random varables, since $T^{-k}$ are i.i.d uniform circle rotations so by the law of large numbers it converges almost surely to the expectation of one of these variables, which is $$ \mathbb E | \mu (A) \cap F^{-1}(B) - \mu(A)\mu(B)|$$ for $F$ a random rotation.

This is nonzero for $A,B$ any two intervals. In fact, it is equal to the long run average in the case that $T$ is an irrational rotation, so this reduces to the case you already know.