What is a canonical term of $\text{Id}_A(x,y)$ if $x$ is not jugdmentally identical to $y$?
Yes, in general $\mathrm{Id}_{A}(x, y)$ will not have a canonical form. Consider the case when $x$ and $y$ are distinct free variables -- obviously you can postulate that $x$ and $y$ are equal, but you can't provide a proof of it.
Even more simply, consider the empty type $0$. It has no canonical forms at all, but there's nothing stopping you from assuming that a variable has type $0$.

For standard type theory, it is provably the case that if you have a closed term $t$ of type $\mathrm{Id}_{A}(x, y)$, then $x$ and $y$ are definitionally equal.
Since the language is normalizing and $t$ is well-typed in the empty context, then we know that $t$ will evaluate to $\mathrm{refl}(z)$ for some $z$, and then by inversion we will be able to establish that  $z \equiv x \equiv y$ judgementally.