The conjecture seems to fail for $n=3$ and $(q_1,q_2,q_3) = (9,10,10)$, 
when ${\rm vol}(E) = (2^3/3!) 9 \cdot 10 \cdot 10 = 1200$ but
$ \#(E \cap {\bf Z}^3) = 1199$.  In general it seems (and is probably
not hard to prove) that if $(q_1,q_2,q_3) = (k,k+1,k+1)$ then
$$
\#(E \cap {\bf Z}^3) - {\rm vol}(E) = -\frac23 k^2 + \frac{16}{3} k + 5
$$
which is negative for $k \geq 9$, and increasingly so as $k$ grows.
Other examples with $n=3$ and $q_1 \leq q_2 \leq q_3 \leq 20$ are
$(6,17,18)$, $(7,13,14)$, $(8,15,16)$, $(9,17,18)$, $(9,19,19)$, $(10,19,20)$
with a shortfall of $-1$, $-17/3$, $-17$, $-31$, $-1$, $-143/3$ respectively,
and suggesting similar generalizations.