Is trace invariant under cyclic permutation with rectangular matrices?

Yes, it holds true. Let $A$ be a $n\times m$ and $B$ be a $m \times n$ matrix over the commutative ring $R$, we have \begin{align*} \mathrm{tr}(AB) &= \sum_{i=1}^n (AB)_{ii}\\ &=\sum_{i=1}^n \sum_{j=1}^m A_{ij}B_{ji}\\ &= \sum_{j=1}^m \sum_{i=1}^n B_{ji}A_{ij}\\ &= \sum_{j=1}^m (BA)_{jj}\\ &= \mathrm{tr}(BA) \end{align*} So you can just prove the formula by computing.

Yes, the cyclic invariance holds irrespective of the dimensions of the matrices. The trace of a product in either order is simply the sum of all products of corresponding entries.