If Q1 and Q2 are countably enumerable, then is Q1\Q2 countably enumerable?
First of all, notice that not all c.e. sets have a c.e. complement, i.e. there are c.e. sets like $Q$ such that $Q^c=\mathbb{N}-Q$ is not c.e. In fact the  Halting Problem $H$ is such a c.e. set.

To see why $H^c$ is not c.e., notice that if both a set and its complement are semi-decidable, then both are also decidable. And since $H$ is not decidable, then $H^c$ should not be semi-decidable either.

Now for the counter example to your question, let $Q_1 = \mathbb{N}$ and $Q_2 = H$. Then $\mathbb{N} - H = H^c$, is not c.e.


Edit: Explained $H^c$ and added a simpler and more straightforward counter example.