No.

If $D\neq\mathbb{C}$ is a s.c. domain, then a conformal map to a disc extends analytically beyond the boundary if and only if D is bounded by an analytic Jordan curve.

So let $D\subset\Omega$ be a domain compactly contained in $\Omega$ with non-analytic boundary, let $\phi$ be a conformal isomorphism to the unit disc and let $\gamma$ be the preimage of a round circle (say centred at zero and of radius less than one) under $\phi$.

Any map f as in your question would then agree with $\phi$ up to Möbius postcomposition. This would imply that $\phi$ extends beyond the boundary of D, a contradiction.

If $G\neq\mathbb{C}$ is a s.c. domain, then a conformal map to a disc extends analytically beyond the boundary if and only if $G$ is bounded by an analytic Jordan curve. In particular, if you were to let your domain $\Omega$ depend on the curve $\gamma$, then the answer would be positive.

However, as stated, the answer is negative. Indeed, let $\Omega$ be a simply-connected domain; we shall construct an analytic curve $\gamma$ such that no conformal isomorphism of the interior of $\gamma$ to a disc extends to all of $\Omega$.

Let $G$ be any bounded simply-connected domain whose closure is contained in $\Omega$, such that $G$ has non-analytic boundary. Let $\phi$ be a conformal isomorphism to the unit disc $\mathbb{D}$ and let $\gamma$ be the preimage of a round circle under $\phi$; say $$\gamma := \phi^{-1}( \partial B(0,1/2) )$$ (where $B(z,\delta)$ is the ball of radius $\delta$ centred at $z$).

Suppose $f:\Omega \to U\subset\mathbb{C}$ was a conformal isomorphism with $f(\gamma)=\partial\mathbb{D}$. Since a conformal isomorphism from the inside of $\gamma$ to a disc is unique up to post-composition by a Möbius transformation, it follows that there is Möbius $M$ such that $\phi=f\circ M$. But then $\phi$ extends analytically beyond the boundary of $G$, a contradiction.

(Edited as requested to provide further details and adjust notation slightly.)