Does a turing machine halt with input left on tape?
When a Turing machine enters an accepting state, it immediately stops running and accepts whatever the original input string was (regardless of the state of the tape). So given the input aabb, the Turing machine will read the two as and enter the accepting state $q_2$, at which point it will halt and accept aabb. 

In this case it doesn't matter that there's no transition for reading a b from $q_2$ because we've already accepted (though you are correct in general that whenever we're running a Turing machine and we arrive at a state with no transition defined for the character we're reading, we implicitly go to a dead state and reject).