In terms of the question "do there exist a slightly modified free-return trajectory..." I don't know (I know that's a weird way to start an answer, but bear with me). Getting to Lunar Polar orbit is not made any easier by starting in Earth polar orbit, that is overkill.
You basically need to fly over the poles of the moon. Given the moon's size and distance, this means that it can be achieved with a relatively low inclination Earth orbit, something probably less than a few degrees inclined with respect to the moon's orbit, probably lower. It's all about angles - the moon's angular size is small, and the distance is large, so if you are inclined just a little bit, then over the distance to the moon your orbit will go over the poles.
Performing a plane change in orbit is prohibitively expensive no matter which body you are orbiting. To change just a few degrees in low Earth orbit would take several hundred m/s of delta v. To change from equatorial to polar in a lunar orbit would be insane (I can't find a good number, but there is an equation here. You'll probably find that the vast majority of the delta-v budget of a mission would be taken up by the plane change.
NASA has studied such orbits extensively, since mapping satellites are best placed in such orbits in order to get as much surface coverage as possible. Here is a NASA trajectory study of this very thing - low energy transfers.
Whether a free-return polar trajectory exists or not is an interesting question. I strongly suspect not, but I'm happy to be proven wrong here. Free-return in the sense of return to the surface without expending fuel would require a small gravitational assist from the moon, which is not possible for a purely polar orbit. Perhaps a slightly less-than-polar orbit, where some small in-plane velocity change can occur, is possible, so don't quote me on the impossibility of such an orbit yet.