What you want is not possible. The basic idea is that it is possible for a diffeo $\Omega_1\to \Omega_2$ that extends to a homeo $\bar{\Omega}_1 \to \bar{\Omega}_2$ to "almost fold up" $\partial\Omega_1$.
To give an explicit example: let $\Omega_1$ be the upper half-unit-disk, in polar coordinates the set $\{0< r < 1, 0 < \theta < \pi\}$.
Let $f(r,\theta) = (r,(2 - r^q)\theta)$ for some $q \gg 1$. Then $\Omega_2 = \{0 < r < 1, 0 < \theta < (2-r^q) \pi \}$. Let $g$ be the inverse mapping, we find $g(s,\phi) = (s, (2 - s^q)^{-1} \phi)$. Both $f$ and $g$ are clearly smooth, and extend continuously to the boundary of the domains $\Omega_1$ and $\Omega_2$ respectively.
Furthermore,
$$ |\nabla f|^2 = 1 + q^2 r^{2q} \theta^2 + (2-r^q)^2 $$
and
$$ |\nabla g|^2 = 1 + q^2 s^{2q} \phi^2 (2 - s^q)^{-4} + (2 - s^q)^{-2} $$
are uniformly bounded on their domains. So we have $W^{1,p}$ automatically.
Now consider the two points $(r,0)$ and $(r,\pi)\in \bar{\Omega}_1$, these correspond to $(r,0)$ and $(r, 2\pi - r^q\pi)\in \bar{\Omega}_2$. In the domain their distance is $2r$. In $\bar{\Omega}_2$ their distance is on the order of $r^{1+q}$. This shows that $g$ is not in $C^{0,\alpha}$ for any $\alpha > \frac{1}{1+q}$.
Replacing $r^q$ by a function vanishing to infinite order at $0$ will get you then an example that is not in $C^{0,\alpha}$ for any $\alpha > 0$.
(Note that this is firmly an issue related to the fact that $\Omega_2$ is not Lipschitz, and then fact that we are measuring Hoelder-ness using the ambient metric on $\mathbb{R}^2$.)
