Perfect matchings in certain classes of hypergraphs

The answer is no. That is, already for $n=3$, $r = 3$ and $k = 2$ there is an $r$-partite $r$-uniform $k$-regular hypergraph that doesn't contain a perfect matching.

Let $v_1,v_2,v_3$ be the first part, $u_1,u_2,u_3$ be the second part and $w_1,w_2,w_3$ the third. Let the hyperedges be $$(v_1,u_1,w_1),(v_1,u_2,w_2),(v_2,u_2,w_1),(v_2,u_3,w_3),(v_3,u_3,w_3),(v_3,u_1,w_2) .$$

It is easy to verify that the resulting hypergraph is 2-regular. However, there is no perfect matching. To see this, consider $w_1$. It belongs to the first and third edge. If we take the first, then we can't take the second (because both contain $v_1$) and so we must take the third edge if we want to cover $u_2$. However, this is impossible because then $w_1$ is covered twice. Hence there is no perfect matching.