Okay, a further question. Despite its simplicity, it is not homework.

Atiyah and Macdonald (Ch. 2, Ex. 19.iv))
claim that if $M$ is a sum of submodules $M_i$,
then $\text{Supp}(M) = \bigcup \text{Supp}(M_i)$,
not specifying what size of set the indices $i$ are to run over.
I assumed initially the index set was to be assumed arbitrary,
and came up with an argument, which it now occurs to me may be wrong,
as your example makes me wonder if the sum should really be finite.
(Or I'm possibly just (continuing to be) stupid.)

Continue with your example,
so $M = \mathbb Q$ is a $A =\mathbb Z$-module,
and $B = \mathbb Z / p \mathbb Z$.
Now $\mathbb Q$ is generated over $\mathbb Z$
by the elements $\frac 1 n \in \mathbb Z$,
so it is a sum of the submodules $\frac 1 n \mathbb Z$,
Thus the elements $z_n = 1 \otimes \frac 1 n$ generate
$B \otimes_A M = \mathbb Z / p \mathbb Z \ \otimes_{\mathbb Z} \mathbb Q$;
in fact all are zero, since the generator
$z_n = p \otimes \frac 1 {pn}$
is zero already in
$\mathbb Z / p \mathbb Z \ \otimes_{\mathbb Z} \frac{1}{pn}\mathbb Z$.

So $\mathbb Z / p \mathbb Z \ \otimes_{\mathbb Z} \mathbb Q = 0$
has empty support,
even though the $\mathbb Z / p \mathbb Z$-submodules
$\mathbb Z / p \mathbb Z \ \otimes_{\mathbb Z} \frac 1{n} \mathbb Z
\cong \frac 1{n} \mathbb Z / \frac p{n} \mathbb Z \cong \mathbb Z / p \mathbb Z$
all have $(0)$ in their support.

Is this right, or am I making an obvious mistake?
Should the sum of modules be assumed finite for the proposition to hold?