The trace operator maps $H^1(\Omega)$ into $H^{1/2}(\partial \Omega)$, I have used this fact several times, and know of references where to find the proof. Let us assume that the boundaries are arbitrarily smooth. The trace operator then maps $C^1(\Omega)$ into $C^1(\partial \Omega)$.
Do you have an intuitive reason why we lose regularity in the weak case?
I think this may be a good way to get valuable understanding of how weak and strong derivatives differ.
The difference in regularity here is less about "weak derivatives vs strong derivatives", and more about $L^2$-based vs $L^\infty$-based spaces.
In general, the trace operator on suitable defined Sobolev spaces maps $$\mathcal{T}:W^{s,p}(\Omega)\to W^{s-1/p,p}(\partial \Omega).$$
From this we see that there is less loss the higher $p$ is, but the classical spaces $C^1$ are defined by a sup-norm estimate, which is a classical analogue to $p=\infty$.