Let $S$ be a surface, and let $g$ be a smooth Riemannian metric on $S$.
Let $p \in S$, and set $r(x)=d_g(p,x)$. Suppose that the curvature of $S$ at a point $q \in S$ depends only on $r(q)$, at least for points sufficiently close to $p$.
Does there always exist a coordinate $\theta$ defined on some punctured ball $U=B_r(p) \setminus\{p\}$ such that on $U$, $g$ is given by $$g=dr^2+q^2(r)d\theta^2. \tag{1}$$
I am quite certain we can always "complete" $r$ to an orthogonal coordinate system $(r,\theta)$, i.e. find $\theta$ such that $$g=dr^2+q^2(r,\theta)d\theta^2. \tag{2}$$ A computation gives $$ K(r,\theta)=-\frac{q_{rr}}{q}, $$ which should be independent of $\theta$. ($K$ is the curvature of $S$).
Finding a coordinate $\tilde \theta$ and $\tilde q(r)$ such that $$g=dr^2+\tilde q^2(r)d\tilde \theta^2, $$ is equivalent to solving $$ q(r,\theta)d\theta=\tilde q(r)d\tilde \theta. $$ Taking the exterior derivative, we get $$ q_r dr \wedge d\theta=\frac{\tilde q_r q}{\tilde q}dr \wedge d\theta, $$ so $$ q_r=\frac{\tilde q_r q}{\tilde q}=F(r)q, \tag{3} $$ where $F(r):=\frac{\tilde q_r}{\tilde q}$.
Differentiating this, we get $$ q_{rr}=F'(r)q+F(r)q_r, $$ so finally $$ -K(r)=\frac{q_{rr}}{q}=F'(r)+F^2(r). \tag{4} $$
So, to construct such $\tilde q$, we first need to find a solution $F$ to the Riccati type equation $(4)$, and then check whether equation $(3)$ holds. This seems to be only the starting point...