The following implication is true (as you mention):
$$q(X) \in \Bbb Z[X] \text{ monic reducible } \implies \bar q(X) \in \Bbb Z/p\Bbb Z[X] \text{ reducible for all prime } p$$
The contrapositive of this implication is:
$$\bar q(X) \in \Bbb Z/p\Bbb Z[X] \text{ irreducible for some prime } p \implies q(X) \in \Bbb Z[X] \text{ irreducible (or non monic) }$$
However, the implication
$$q(X) \in \Bbb Z[X] \text{ irreducible monic } \implies \bar q(X) \in \Bbb Z/p\Bbb Z[X] \text{ irreducible for some prime } p$$
is the converse of the first implication, which doesn't hold in general (as you provide a link to give a counter-example).
The first implication can be proven as follows:
if you assume $q= fg$ with $f,g \in \Bbb Z[X] \setminus \{±1\}$, then $\bar q = \bar f \bar g \in \Bbb Z/p\Bbb Z[X]$. If you suppose that the dominant coefficient of $q$ is not a multiple of $p$ (e.g. $q$ monic), then both $\bar f$ and $\bar g$ have positive degree, hence $\bar q $ is reducible.
More generally, if $A$ is a UFD, $a \in A$ is an element, $f \in A[X]$ is a polynomial with content $1$ and with dominant coefficient invertible in $A/(a)$, then the irreducibility of $\bar f$ in $A/(a)[X]$ gives the irreducibility of $f$ in $A[X]$. [Berhuy, theorem IV.2.3].