It seems that one can use a similar trick to that used in your construction on $K_{2^k}$ for any $K_{4n}$: label the vertices by elements of $G = (\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2n\mathbb{Z})$ and assign edge colors by the difference of the two endpoints (EDIT: as noted in Ilhee's comment, this won't be well-defined; am thinking about possible modifications). Since the sum of all the elements of $G$ is the identity, the same argument which shows your coloring of $K_{2^k}$ is rainbow-free applies to this coloring of $K_{4n}$.
Unfortunately this argument fails when we try to use it to label $K_{4n+2}$ by elements of $(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/(2n+1)\mathbb{Z})$ since the sum of the elements of the group is not the identity in this case.
My gut tells me there is some clever way to label the vertices in the $K_{4n+2}$ so as to define colors on the edges which force your condition, but I don't see it; I'll have to think about it some more.
