The axiom says: every nonnempty set bounded above has a supremum. In the case of $\{ q \in \mathbb{Q} \mid q^2\le2\}$ we do not have a supremum in $\mathbb{Q}$, but the one in $\mathbb{R}$.
Is it necessary for the supremum to be in the same set? If so, why? The set I wrote here will be still bounded by $\sqrt{2}$, won't it?
On
It makes no sense to say only that a set $S$ has a supremum. What makes sense is to say that $S$ has a supremum in a specified universe $U$. In other words, having a supremum is not simply a property of a set but a property of a pair of sets $S$ and $U$. As you observed, one and the same $S$ (like $\{q\in\mathbb Q:q^2<2\}$) might have a supremum in one universe (like $\mathbb R$) and not have a supremum in another universe (like $\mathbb Q$).
(Admittedly, this is essentially the same answer that José Carlos Santos already gave, but I hope that, by making it a little longer, I've also made it easier to understand.)
In $\mathbb Q$, the set $\{q\in\mathbb Q\,|\,q^2\leqslant2\}$ has no supremum. In $\mathbb R$, it has one (which is $\sqrt2$).