I would argue that
$$\int_{{SU}(N)}s_\lambda(u)\overline{s_\mu(u)}du=\int_{{U}(N)}s_\lambda(u)\overline{s_\mu(u)}du=\delta_{\lambda\mu}\;\;\text{if}\;\;\min(|\lambda|,|\mu|)<N.\qquad\qquad(\ast)$$
Starting from
$$\int_{{U}(N)}f(u,\bar{u})du=\int_0^{2\pi}\frac{d\phi}{2\pi}\int_{{SU}(N)}f(e^{i\phi}u,e^{-i\phi}\bar{u}) du$$
the equality $(\ast)$ follows if $|\lambda|=|\mu|$, because $s_\lambda(e^{i\phi}u)=e^{i|\lambda|\phi}s_\lambda(u)$.
$s_\lambda(u)$ is a symmetric homogeneous polynomial of degree |λ| in the eigenvalues of $u$.
If $|\lambda|\neq|\mu|$ and $\min(|\lambda|,|\mu|)<N$ we have sufficient freedom to change the sign of the integrand by inverting eigenvalues without changing $\det u$, so the integral over ${SU}(N)$ vanishes by symmetry. (I need to make this part of the argument more precise.)