No. Let $\gamma$ be the image of $S^1$ under $z\mapsto e^z$. Any conformal isomorphism $g$ from $D$ to the region bounded by $\gamma$ must have the form $z\mapsto e^{R(z)}$, where $R$ is some degree one rational function (such that $R(D)=D$). Let $\Omega$ be ny simply connected region having both $\gamma$ and $0$ in its interior. A conformal map $f$ from $\Omega$ to a region containing $D$ such that $f(\gamma)=S^1$ would have to have some such map $g$ as its inverse; but $0$ is never in the image of $z\mapsto e^{R(z)}$.

No. Let $\gamma$ be the image of $S^1$ under $z\mapsto e^z$. Any conformal isomorphism $g$ from $D$ to the region bounded by $\gamma$ must have the form $z\mapsto e^{R(z)}$, where $R$ is some degree one rational function (such that $R(D)=D$). Let $\Omega$ be any simply connected region having both $\gamma$ and $0$ in its interior. A conformal map $f$ from $\Omega$ to a region containing $D$ such that $f(\gamma)=S^1$ would have to have some such map $g$ as its inverse; but $0$ is never in the image of $z\mapsto e^{R(z)}$.