Fourier transforms wouldn't help you with a rotation. You'd just end up having to rotate the matrix of Fourier coefficients, instead of rotating the original image.
Consider for example an image made of a perfect sine wave along the x-axis with wave-vector $(k, 0)$. (The wave-vector is the spacial frequencies along the $x$ and $y$ axes). The Fourier transform of this would be all black, with a single white pixel at position $(k, 0)$.
When you rotate the image by an angle $\theta$, you'll get a perfect sine wave along an oblique axis. If you draw a diagram and do a little trigonometry, you can see that the new wave-vector will be $(k \cos \theta, k \sin \theta)$. (It helps for this part to know that the frequency $k$ is one over the period.)
Therefore, the Fourier transform of the rotated result would be all black, with a single white pixel at position $(k \cos \theta, k \sin \theta)$. But that's just what you'd get if you rotated the Fourier transform of the original wave by $\theta$.
As for the Gibbs phenomenon, it's an issue when downsampling an image, or compressing it by frequency quantization (as in JPEG). But simply taking the Fourier transform of an image doesn't introduce any Gibbs ringing. The Fourier transform is lossless and perfectly reversible if done properly: it represents all the information in the original image.