Let $y$ and $z$ be two different integer roots of $q$. Then $y^3 + a y^2 + b y + c = 0$ and $z^3 + a z^2 + b z + c = 0$. We know that $p$ divides $a$, $b$ and $c$. Since $y^3 = -c - b y - a y^2$, $p$ divides $y^3$. Similarly, we show that $p$ divides $z^3$. As $p$ is a prime, it must divide $y$ and $z$.

By subtracting the two equalities above we get $y^3 - z^3 + a(y^2 - z^2) + b(y - z) = 0$ or
$$(y - z)(y^2 + y z + z^2 + a(y + z) + b) = 0.$$
Since $z \neq y$, this implies $y^2 + y z + z^2 + a(y + z) + b = 0$. Now, $p$ is a prime and it divides $y$, $z$ and $a$, so $p^2$ divides $y^2 + y z + z^2 + a(y + z) = -b$, hence $p^2$ divides $b$.

We know that $c = -y^3 - a y^2 - b y$. Since $p$ divides $y$, $p^2$ divides $y^2$ and $p^3$ divides $y^3$. Since $p$ also divides $a$ and $p^2$ divides $b$, we can conclude that $p^3$ divides $c$.