Even with the assumption that $e^{iq}A+e^{-iq}B$ is Hermitian, that is $B=A^* $, the claim is not true. The characteristic polynomial $P(X,z)$ of $zA+\bar zA^*$, written in terms of the real and imaginary parts $z=Y+iZ$ is a Hyperbolic polynomial. Actually, every hyperbolic polynomial can be written that way, according to a conjecture of P. Lax, proved by Helton and Vinnikov. For instance, $$X^2-Y^2-Z^2=\det\begin{pmatrix} X-Y & Z \\\\ Z & X+Y \end{pmatrix}$$ corresponds to the matrix $$A=\begin{pmatrix} 1 & i \\\\ i & -1 \end{pmatrix}.$$ This matrix turns out to be nilpotent, as you require. Perhaps this is the case that you investigated, because here the eigenvalues of $zA+\bar zA^* $ are $\pm|z|$, thus constant $\equiv\pm1$ when $z=e^{iq}$. However, the following is a counter-example: $$A=\begin{pmatrix} 0 & 1 & 1 \\\\ 0 & 0 & 1 \\\\ 0 & 0 & 0 \end{pmatrix}.$$ When $z$ runs over the unit circle, the eigenvalues of $zA+\bar zA^* $ form a cardioid, plus its bitangent.
Edit. The parity of $n$ does not matter. As said by David, there are plenty of hyperbolic polynomials. Take $n=4$ and $P_0(X,z)=(X^2-|z|^2)(X^2-2|z|^2)$. It is strictly hyperbolic (distinct roots for every nonzero $z$). Then every polynomial $P(X,Y,Z)$ close enough to $P_0$ is hyperbolic. According to Lax--Helton--Vinnikov, $P$ is the characteristic polynomial of some $zA+\bar zA^* $. Actually, you can start from the nilpotent matrix $A_0$ attached to $P_0$ and perturb it keeping it nilpotent.. Still you get plenty of $P$, most of them non-constant over the unit circle.