A Noetherian ring is reduced if and only if it is $(R_0)$ and $(S_1)$. (See, for example, http://stacks.math.columbia.edu/tag/031R ) This condition can be used to provide a simple proof that your ring $A/\mathfrak{p}$ is reduced.

A ring $R$ is $(R_k)$ iff $A_P$ is a regular local ring for all primes $P$ of height at most $k$. So $R$ is $(R_0)$ iff $R_P$ is regular for all minimal primes $P$. This is a simple condition and clearly holds for your ring $R = A/{\mathfrak q}$ since it has a unique minimal prime and its localization at that prime is a field.

A ring $R$ is $(S_k)$ iff $PR_P$ contains a regular sequence in $R_P$ of length at least $\min(k,\mbox{ht} P)$ for every prime $P$ of $R$. So $R$ is $(S_1)$ iff $PR_P$ contains a nonzerodivisor in $R_P$ for every non-minimal prime $P$ of $R$. This condition also clearly holds for your ring $R = A/\mathfrak{q}$.

A Noetherian ring is reduced if and only if it is $(R_0)$ and $(S_1)$. (See, for example, http://stacks.math.columbia.edu/tag/031R ) This condition can be used to provide a simple proof that your ring $A/\mathfrak{p}$ is reduced.

A ring $R$ is $(R_k)$ iff $A_P$ is a regular local ring for all primes $P$ of height at most $k$. So $R$ is $(R_0)$ iff $R_P$ is regular for all minimal primes $P$. This is a simple condition and clearly holds for your ring $R = A/{\mathfrak q}$ since it has a unique minimal prime and its localization at that prime is a field. The assumption that $\mathfrak{q}$ is locally principal is not required here.

A ring $R$ is $(S_k)$ iff $PR_P$ contains a regular sequence in $R_P$ of length at least $\min(k,\mbox{ht} P)$ for every prime $P$ of $R$. So $R$ is $(S_1)$ iff $PR_P$ contains a nonzerodivisor in $R_P$ for every non-minimal prime $P$ of $R$. This condition also clearly holds for your ring $R = A/\mathfrak{q}$. In general the condition required beyond $(R_0)$ is precisely the condition $(S_1)$.