There are two cases. In case $p|(q-1)$, then the multiplicative group of $\mathbb F_q$, cyclic of order $q-1$, has a subgroup of order $p$, whose elements are the roots of $\Phi_p$, since these are the primitive $p$-th roots of unity. In this case, all the roots of $\Phi_p$ are in $\mathbb F_q$, and they lift to $\mathbb Z_q$ as $p-1$ distinct elements of the $q$-adic integers. They are all the roots of $\Phi_p$. There can’t be any others, in $\mathbb Q_q$ or any other integral domain containing $\mathbb Z_q$.
The other case is that $p$ doesn’t divide $q-1$, in which case neither $\mathbb F_q$ nor $\mathbb Z_q$ has primitive $p$-th roots of unity. This may be the case that was confusing James. Here, $\Phi_p$ has roots in a proper extension of $\mathbb F_q$, as @Jyrki remarks, and these lift to the corresponding unramified proper extension of $\mathbb Z_q$ but not to $\mathbb Z_q$ itself.
So, in both cases, there are just as many primitive $p$-th roots of unity in $\mathbb Z_q$ as in $\mathbb F_q$.