The answer is also yes if one of the primes, say $r$, is $2$, because then $\zeta_{2r}+\zeta_{2r}^{-1}=0$ and $\zeta_{2q}+\zeta_{2q}^{-1}=\zeta_{2q}(1+\zeta_{2q}^{-2})$ is a unit (as $\zeta_{2q}^{-2}$ is a primitive $q$th root of $1$ and $q$ is an odd prime).
Edit:
(1) Note that if both primes are odd then $\zeta_q+\zeta_q^{-1}+\zeta_r+\zeta_r^{-1}$ is also a unit. Indeed, because when you multiply all terms by $-1$ you get$-\zeta_q$ is a primitive $\zeta_{2q}+\zeta_{2q}^{-1}+\zeta_{2r}+\zeta_{2r}^{-1}$$2q$th root of $1$ (this relies on $q$ being odd), so let's call it $\zeta_{2q}$, and likewise for $r$. But this conclusionThen $-(\zeta_q+\zeta_q^{-1}+\zeta_r+\zeta_r^{-1})=\zeta_{2q}+\zeta_{2q}^{-1}+\zeta_{2r}+\zeta_{2r}^{-1}$, and we know that the latter is falsea unit.
(2) Note also that if one of the primes, say $r$, is $2$ then $\zeta_q+\zeta_q^{-1}+\zeta_r+\zeta_r^{-1}$ is not a unit: it is equal to $\zeta_q+\zeta_q^{-1}-2$, so it is a unit times the square of $\zeta_q-1$, but the latter (unlike $\zeta_q +1$) is not a unit because it goes to $0$ under the unique ring homomorphism $\mathbb Z [\zeta_q]\to \mathbb Z/q$, which takes $\zeta_q$ to $1$.