I believe that if you have a sufficiently regular (say Lipschitz) bounded domain $\Omega\subset\Bbb R^n$, then you can extend a function $u\in H^1_0(\Omega)$ by zero, and the extension lies in $H^1(\Bbb R^n)$.
Consider the unbounded wedge $\Omega=\{(x,y)\in\Bbb R^2:- \infty<x<y,y>0\}$, and the subset $\Omega_R=B_r(0)\cap\Omega$ for some $R>0$ bounded.
Let $\Gamma = \{(x,y)\in\overline{\Omega}_R:x^2+y^2=R^2\}$ (i.e., the full curved boundary) consider the space $H^1_E(\Omega_R)=\{v\in H^1(\Omega_R):v|_{\Gamma}=0\}$.
My question is can I extend a function $u\in H^1_E(\Omega_R)$ by zero to the whole of $\Omega$, and have the extension lie in $H^1(\Omega)$?
If this does or doesn't hold it would be great to hear an argument as to why/ why not.
Since your region has a simple geometry this should be a consequence of the "absolutely continuous on lines" characterization of $H^1$, e.g. section 4.9.2 in Evans and Gariepy.
By the way, the zero extension of a function in $H^1_0(\Omega)$ will belong to $H^1(\mathbb R^n)$ with no restriction whatsoever on the boundary of $\Omega$.