This is only a partial answer, that rests on the non-trivial assumption that there exists a solution to the Cheeger problem. It is an expansion of a proof given in "The isoperimetric problem on surfaces (Howard, Hutchings, Morgan)".
The answer will not be accepted until completed, but if anybody would prove the assumption in a separate answer I will be glad to mark it as accepted.
Let $\gamma$ be the minimizer of the Cheeger problem described in the question, whose existence we assume true. Let $A_\gamma$ be the inscribed surface.
First, we claim that the origin must lie inside $\gamma$: $O \in A_\gamma$. To prove this we proceed by contradiction. We can trivially exclude the cases where $\gamma$ has more than one connected component. Moreover, to avoid pathological cases where the upcoming construction would not work, we replace $\gamma$ by its circular symmetrization, whose Cheeger constant is not worse than the one of $\gamma$. Now, let $l_0$ be a tangent to $\gamma$ passing through $O$, and define $\theta_0$ to be the smallest angle such that $\gamma$ is contained in the cone limited by $l_0$ and its $\theta_0$-rotation. We denote this second line by $l_1$, it also tangent to $\gamma$. If $\theta_0 \leq \pi$ we replace $\gamma$ by itself plus its reflection through the line $l_1$, this is clearly not changing $h(\gamma)$. Now repeat the same construction: the new angle $\theta_1$ is $\theta_1 = 2\theta_0$. We repeat the construction until $\theta_n >\pi$.
This new $\gamma$ is still a solution of the Cheeger problem, indeed this construction does not change the Cheeger constant or the distance from the origin, but is now concave! Since the metric is radially symmetric, and $\theta > \pi$, the geodesic connecting the two extremal points of $\gamma$ is strictly shorter than the shortest path connecting them along $\gamma$. Moreover also the area of $A_\gamma$ increases, yielding a better Cheeger constant. We can thus assume that $O\in A_\gamma$.
Second, we claim that the solution must be a circle, i.e. radially symmetric. Indeed let $l$ be a line passing through the origin $0$ and splitting $\gamma$ into two, not necessarily connected, parts. We replace $\gamma$ by the part with the better Cheeger constant plus its reflection through $l$. Let $m$ be the line through $O$ and orthogonal to $l$, again we replace $\gamma$ by the part with the better Cheeger constant plus its reflection through $m$. Now $\gamma$ is invariant by reflections through $l$ and $m$, thus by $\pi$-rotations. Specifically, every line through the origin splits $\gamma$ in two parts with equal area and perimeter, thus with equal Cheeger constant. If $\gamma$ is not a circle, there exists a line $s$ through the origin that intersects it not orthogonally. Replacing $\gamma$ by the one of the half in the splitting by $s$ plus its reflection does not vary the Cheeger constant, but yields a concave curve. Since the metric is monotone decreasing, the convex hull of this curve has smaller perimeter and larger area, thus a better Cheeger constant. Contradiction.
As a final remark I would like to point out that, in the last step, we are actually using the monotonicity of the metric.