For generic $A,B$ the matrix $A$ is invertible and $B=CA$, where $C$ is also generic. We have $(A\otimes B-B\otimes A)(u\otimes v)=Au\otimes CAv-CAu\otimes Av$. The vectors $Au$ run over the whole $\mathbb{R}^n$, so the image of the operator $A\otimes B-B\otimes A$ is the same as the image of $I\otimes C-C\otimes I$ (where $I$ is the identity matrix). Choosing an eigenbasis for $C$ (it exists if $C$ is generic, possibly it is complex but the rank is the same over $\mathbb{R}$ and over $\mathbb{C}$), where $Ce_i=\lambda_i e_i$, we get $$(I\otimes C-C\otimes I)(e_i\otimes e_j)=(\lambda_j-\lambda_i)e_i\otimes e_j.$$
So, $(e_i\otimes e_j)$ is an eigenbasis, and all but $n$ eigenvalues are non-zero.