whether it's possible to solve the cube in a way that the colors match correctly but the text is misaligned
Yes. But the total amount of misalignment (if I remember correctly; it's been a while since I played with a Rubik's cube) must be a multiple of $\pi$ radians. So if only one face has the center piece misaligned, it must be upside down. On the other hand it is possible to have two center pieces simultaneously off by quarter-turns. (This is similar to the fact that without taking the cube apart, you cannot change the orientation of an edge piece (as opposed to center piece or corner piece) while fixing everything else.)
(I don't actually have a group-theoretic proof for the fact though; this is just from experience.)
Edit: Henning Makholm provides a proof in the comments
Here's the missing group-theoretic argument: Place four marker dots symmetrically on each center, and one marker at the tip of each corner cubie, for a total of 32 markers. A quarter turn permutes the 32 markers in two 4-cycles, which is even. Therefore every possible configuration of the dots is an even permutation away from the solved state. But misaligning one center by 90° while keeping everything else solved would be an odd permutation and is therefore impossible.