As Robert writes, the answer is positive for *round* circles, which follows from the fact that Möbius transformations map circles to circles (and you can map any circle contained in the unit disc to a circle with its centre at 0).

As Neil mentions, the answer is negative for arbitrary curves. Indeed, this is trivial if the curves in question are not analytic. If the curves are analytic, but not round circles, then you *can* extend the conformal map across the boundary to a larger annulus by Schwarz reflection, but not to the whole disc - again, because Möbius transformations map circles to circles, and any conformal isomorphism between discs is Möbius!

Finally, to respond to the new version of the question raised in your comment - Suppose that your annulus has modulus M, and let $\psi$ be a conformal map that takes the interior of $C_2$ to the unit disc, **normalised in such a way that $\psi^{-1}(0)\notin \Omega$.** (This assumption was missing from your question, but without it it is clear that you cannot say anything.)

Consider the annulus $A := 1/\psi(\Omega)$. This is a doubly connected region of modulus M, separating the unit circle from $\infty$. Let $R>1$ be the smallest number such that $A$ omits a point of modulus $R$; wlog this point is $R$ itself. What you are asking for is a lower bound on $R$.

It is well-known that, among all such annuli, the *Grötzsch annulus*, which is the complement of the closed unit disc and the segment $[R,\infty)$, has the largest modulus, which we denote $M(R)$. (See Ahflors, "Conformal Invariants".) Hence we see that $M(R)\geq M$.

In other words, let $\rho>1$ be such that $M(\rho) = M$. (Such $\rho$ exists because $M$ is clearly monotone and continuous, and tends to $0$ and $R\to 1$ and to $\infty$ as $R\to\infty$.) Then $R\geq \rho$, and hence
$$\psi(\Omega) \subset D(0,1/\rho),$$
which answers your question. Clearly this bound is sharp, by construction.

There are explicit estimates for the modulus of the Grötzsch annulus; see e.g. Ahlfors's book cited above.

Another interpreation of your question can be given when your inner boundary $C_1$ is not a round circle, but a **quasicircle**. (This is true whenever the curve is smooth.) In this case, you can extend the conformal isomorphism to the round annulus to a map that will in general not be conformal, but at least be **quasiconformal**, with constants depending on the constants in the quasicircle definition. Quasiconformal maps satisfy some nice properties, so again these will give you some control over the image of your curve.