There is a very good theory of the boundary behavior for Riemann maps, the Caratheodory theory of ends. Riemann maps for an open set extend continuously to the boundary of a disk provided the frontier is locally connected.
I assume your definition is in terms of the Riemann maps that fix the north or south pole and have derivative a positive real number at those points. If so, then Riemann maps depend continuously in the $L^\infty$ metric on Jordan curves as you have described, so this gives a metric.
Is the metric complete? A sequence of the pair (lower hemisphere map, upper hemisphere map) that is Cauchy in the uniform ($L^\infty$) topology converges to a pair of continuous maps that agree on the equator, so the glued maps give a continuous image of the sphere that is a homeomorphism at least in the complement of the equator. If it is not injective, the preimages of points would need to be intervals, otherwise the topology would be destroyed.
But that's impossible. If you push forward the measure $ds$ on the equator by a Riemann map, it never has atoms. It is the same as hitting measure for Brownian motion in the image: if you start at the north pole and follow a Brownian path, where does it first arrive at the boundary? This is always a diffuse measure.
So, you're right: it's a complete metric on the set of Jordan curves you described.
Note. Any quasisymmetric map (I won't define it here, but Holder is sufficient) from the circle to the circle arises from a pair of Riemann maps for a Jordan curve, although not necessarily in the annulus you described, and the Jordan curve in this case is known as a quasi-circle. However, there is no known good characterization of which gluing maps from the circle to the circle give Jordan curves in general. Continuity is not sufficient: there are counterexamples using things that are locally (for example) the graph of $\sin(1/x)$ plus the interval $[-1,1]$ on the $y$-axis, where the gluing map for the Riemann maps to the two sides extends continuously across the discontinuity of the graph. Better characterizations of what gluing maps give what topology and geometry are very hard, but of great interest in complex dynamics and some other subjects.
Here is a little progress towards AC.
Theorem.
ICF implies the dual Cantor-Schröder-Bernstein
theorem, that is $X$ surjects onto $Y$ and $Y$ surjects onto $X$,
then they are bijective.
Proof. You explain this in the edit to the question. If
$X\twoheadrightarrow Y$, then $2^Y\leq 2^X$ by taking pre-images,
and so if also $Y\twoheadrightarrow X$, then $2^X\leq 2^Y$ and so
$X\sim Y$ by ICF. QED
Theorem. ICF implies that there are no infinite D-finite
sets.
Proof. (This is a solution to the exercise that you mention.) If $A$ is infinite and
Dedekind-finite, then let $B$ be the set of all finite
non-repeating finite sequences from $A$. This is also D-finite,
since a countably infinite subset of $B$ easily gives rise to a
countably infinite subset of $A$. But meanwhile, $B$ surjects onto
$B+1$, since we can map the empty sequence to the new point, and
apply the shift map to chop off the first element of any sequence.
So $B$ and $B+1$ surject onto each other, and so by the dual
Cantor-Schöder-Bernstein result, they are bijective,
contradicting the fact that $B$ is D-finite. QED
Here is the new part:
Theorem. ICF implies that $\kappa^+$ injects into
$2^\kappa$ for every ordinal $\kappa$.
Proof. We may assume $\kappa$ is infinite. Notice that
$2^{\kappa^2}$ surjects onto $\kappa^+$, since every
$\alpha<\kappa$ is coded by a relation on $\kappa$. Since
$\kappa^2\sim\kappa$, this means
$2^\kappa\twoheadrightarrow\kappa^+$ and consequently
$2^{\kappa^+}\leq 2^{2^\kappa}$, by taking pre-images. It follows that
$2^{2^\kappa}=2^{2^\kappa}\cdot 2^{\kappa^+}=2^{2^\kappa+\kappa^+}$ and
so by ICF we get $2^\kappa+\kappa^+=2^\kappa$, which implies
$\kappa^+\leq 2^\kappa$, as desired. QED
This conclusion already contradicts AD, for example, since AD
implies that there is no $\omega_1$ sequence of distinct reals,
which violates the conclusion when $\kappa=\omega$. In particular,
this shows that ICF implies $\neg$AD, and so in every AD model, there are sets of different cardinalities, whose power sets are equinumerous.
Best Answer
It is possible to change your argument so that the choice is over countable set; hope this is good enough. Namely, topology on the plane has countable base (say, circles at rational points with rational radii); let's index this base as $U_1,\dots,U_n,\dots$; your argument can be used to construct a sequence of Jordan curves $C_1,\dots,C_n,\dots$ such that $C_{i+1}$ is contained in the interior of $C_i$, and $U_i$ is not contained in the interior of $C_i$.