You can always find a map $\Phi$ (unique up to shifts) such that $\Im\mathfrak{m}(\Phi(z)-z)$ is bounded, but, in general, $\Re\mathfrak{e}(\Phi(z)-z)$ may be unbounded.

Let $h$ be the bounded harmonic function in $\Omega$ satisfying the boundary conditions $h(z)=\Im\mathfrak{m} z$, $z\in\partial\Omega$ (such a function can be constructed, e. g., as the exit expectation of the Brownian motion). The function $g(z)=\Im\mathfrak{m}z-h(z)$ is zero on the boundary, and, by maximum principle, is negative. Therefore, if $\varphi$ is a conformal map from the unit disc to $\Omega$, then $g(\varphi(z))$ is a multiple of the Poisson kernel in the disc. From this, you can see that if $\tilde{g}$ denotes the harmonic conjugate of $g$, then $\tilde{g}|_{\partial\Omega}$ is increasing and tends to $\pm \infty$ as $\Re\mathfrak{e}z\to\pm\infty$. The argument principle shows that $\tilde{g}+ig$ is a conformal map.

Observe, however, that if $f$ is a smoothed step function, then the normal derivative of $h$ (which is also the tangential derivative of $\tilde{h}$) behaves as $\Re\mathfrak{e}(const/z)$ as $\Re\mathfrak{e}z$ tends to infinity, so $\tilde{h}$ grows logarithmically. This is the worst possible behaviour.