This puzzle is definitely impossible. In fact it seems to be some sort of parity issue, although I'm having trouble with the actual proof. Basically if you always keep the start in the top left corner, then you can only complete a rook's tour if the end square has an odd manhattan distance (x distance + y distance). If you put the end square an even manhattan distance away, then you create this issue where you're always unable to hit one square.
For example, even parity, impossible:



Odd parity, possible:




