The answer to the specific question is yes. A good way to understand this question is to determine the metric under geodesic polar coordinates as explicitly as possible. This is the course I pursued in the original answer, but a moderator informed me it was too long for MSE. They kindly suggested I port the original answer to a separate personal webpage and shorten this answer. You can see the original answer here. I go into far more detail and provide derivations / proofs for the various lemmas. I also prove some similar characterizations of constant curvature surfaces.
(I should add this text from the moderator's message, "Please note that we also do not encourage users to continually provide answers whose completeness depends on links to external webpages or personal blogs, and that the suggestion here is intended to provide a compromise for this "one time occurrence".")
First let's show that for every point $p \in S$, if all concentric geodesic circles within a geodesic ball centered at $p$ have constant geodesic curvature, then the curvature is a function only of the distance to $p$. This can essentially be found in a 1901 paper [1] by J. K. Whittemore.
Consider an orthonormal basis $e_1, e_2$ of $T_pS$. $$x(\rho, \theta) := \exp_p(\phi (\rho, \theta)) := \exp_p(\rho \cos(\theta)\ e_1 + \rho \sin(\theta)\ e_2)$$ is a geodesic polar mapping, where $\exp_p$ is the exponential map at $p$. Note that the pushforward $\phi_\ast \frac{\partial}{\partial \theta}$ of $\frac{\partial}{\partial \theta}$ is well-defined and equals $0$ when $\rho = 0$. $\phi_\ast \frac{\partial}{\partial \rho}$ is only well-defined when $\rho > 0$, however. Hence the same holds for $x_\ast \frac{\partial}{\partial \theta}$ and $x_\ast \frac{\partial}{\partial \rho}$ for small enough $\rho$, as $\exp_p$ is a diffeomorphism on some neighborhood of the origin.
We define metric coefficients $E = E(\rho, \theta)$, $F = F(\rho, \theta)$, $G = G(\rho, \theta)$ by
\begin{align*}
\begin{bmatrix}
E & F \\
F & G
\end{bmatrix}
:=
\begin{bmatrix}
g\left(x_\ast\frac{\partial}{\partial \rho}, x_\ast\frac{\partial}{\partial \rho}\right) & g\left(x_\ast\frac{\partial}{\partial \rho}, x_\ast\frac{\partial}{\partial \theta}\right) \\
g\left(x_\ast\frac{\partial}{\partial \theta}, x_\ast\frac{\partial}{\partial \rho}\right) & g\left(x_\ast\frac{\partial}{\partial \theta}, x_\ast\frac{\partial}{\partial \theta}\right)
\end{bmatrix}.
\end{align*}
Lemma 8.1.6 of [2] states $E = 1$, $F = 0$, and $G > 0$, except on $(0, \theta)$ for any $\theta$, where $G = 0$. Notably, $E = 1$ means that the radial geodesics are unit-speed curves, and
$F = 0$ means that the geodesic circles are orthogonal to the radial geodesics; this is called Gauss's Lemma, and I prove these in the original answer. These facts show $J_{x(\theta, \rho_0)} = -x_{\ast} \frac{\partial}{\partial \rho}$.
The velocity vector field $\nu$ of a unit-speed reparameterized $\theta$ curve is the restriction of $\frac{x_\ast \frac{\partial}{\partial \theta}}{\sqrt{G}}$ to the curve, so $D_\theta\nu = \nabla_{\frac{x_\ast \frac{\partial}{\partial \theta}}{\sqrt{G}}}\frac{x_\ast \frac{\partial}{\partial \theta}}{\sqrt{G}}$ for fixed $\rho > 0$. From this you can show (see original answer) that the signed geodesic curvature along the reparameterized $\theta$ curve is
\begin{align*}
\kappa = \frac{(\sqrt{G})_\rho}{\sqrt{G}}.
\end{align*}
This is a special case of Liouville's formula for signed geodesic curvature. See [3] for Liouville's formula.
For varying $\rho$ and fixed $\theta$, this gives the first order linear equation $(\sqrt{G})_\rho(\rho, \theta) - \kappa(\rho) \sqrt{G}(\rho, \theta) = 0$. The main assumption of the problem is that $\kappa$ is independent of $\theta$, assuming $\rho$ is small enough. Hence the solution of this equation is given by $\sqrt{G} = e^{U + V}$, for some antiderivative $U = U(\rho)$ of $\kappa$ and some function $V = V(\theta)$. In Theorem 8.3.3 of [2] O'Neill shows $\lim_{\rho \to 0} (\sqrt{G})_\rho(\rho, \theta) = 1$. An application of the Mean Value Theorem further shows that $(\sqrt{G})_\rho(0, \theta) = 1$. I prove these in the original answer.
Our solution for $\sqrt{G}$, together with $(\sqrt{G})_\rho(0, \theta) = 1$, implies $$1 = U'(0)e^{U(0) + V(\theta)},$$
which implies that $V$ is a constant function, since this holds for all $\theta$; we needed the fact that $U$ is independent of $\theta$ to conclude that $V$ is constant. Therefore $G$ is a function only of $\rho$. Because the curvature depends on the metric, we could conclude now that the curvature depends only on $\rho$. Instead of leaving it that, let's give a precise formula for $K$ in geodesic polar coordinates. This will eventually lead to useful formulas for $\sqrt{G}$, and hence for the metric in a geodesic ball.
Consider the Riemann curvature endomorphism,
$$R(X, Y)Z := \nabla_{X}\nabla_{Y} Z - \nabla_{Y}\nabla_{X} Z - \nabla_{[X, Y]}Z.$$
The sectional curvature at a point $p$ of a two dimensional subspace $W$ of $T_pM$ on a Riemannian manifold $M$ is
$$K(W) = K(X, Y) := \frac{g(R(X, Y)X, Y)}{g(X, X) g(Y, Y) - g(X, Y)^2},$$
where $\{X, Y\}$ is any frame which spans $W$ at $p$. (This is the version do Carmo uses in [4]. Lee [0] swaps the second $X$ and $Y$ in the numerator.) To understand this formula, intuitively the numerator is like an obfuscated second fundamental form determinant. Since there is only one two dimensional subspace of $T_pS$, the formula given determines a smooth function $K$ on the surface, called the curvature.
You can show (possibly by reading my original answer) that
\begin{align*}
K = -\frac{(\sqrt{G})_{\rho\rho}}{\sqrt{G}}.
\end{align*}
We therefore have the following second order differential equation for $\sqrt{G}$,
$$(\sqrt{G})_{\rho\rho} + K\sqrt{G} = 0.$$
Under the assumption that small enough geodesic circles have constant geodesic curvature, we have shown that $\sqrt{G}$ is a function only of $\rho$. Therefore we can conclude that $K = -\frac{\sqrt{G})_{\rho\rho}}{\sqrt{G}}$ is a function only of $\rho$. Showing this was our first goal, and with this we are almost done.
Let $p \in S$ and consider a geodesic ball $B$ such that all concentric circles of $p$ within $B$ have constant geodesic curvature. Let $q \in B$, so that $q$ is on the boundary of one of the concentric circles $C$ of $p$ which is contained in $B$. By assumption, $q$ has a geodesic ball $D$ such that all concentric circles of $q$ within $D$ have constant geodesic curvature. The main fact we have proven shows that the curvature of $S$ is constant along $C$. All small enough concentric geodesic circles of $q$ intersect $C$. This is because $\exp_q$ is a diffeomorphism locally and the same
happens for the local preimage of the smooth curve $C$ by $\exp_q$. Therefore all points on small enough concentric circles of $q$ have the same curvature as the points on $C$, and so the curvature is constant in a neighborhood of $q$. This argument applies to every point on $C$, so $C$ has a neighborhood with constant curvature. Taking the union of such neighborhoods over all concentric circles of $p$ smaller than $C$ shows that $p$ has a neighborhood with constant curvature. Finally, connectivity of $S$ shows that $S$ has constant curvature.
Bibliography
[0] Lee, John M., Introduction to Riemannian manifolds, Graduate Texts in Mathematics 176. Cham: Springer (ISBN 978-3-319-91754-2/hbk; 978-3-319-91755-9/ebook). xiii, 437 p. (2018). ZBL1409.53001.
[1] Whittemore, J. K., A note on geodesic circles., Annals of Math. (2) 3, 21-24 (1901). ZBL32.0615.02.
[2] O’Neill, Barrett, Elementary differential geometry (Rev. 2nd ed.)., Amsterdam: Elsevier/Academic Press (ISBN 978-0-12-088735-4/hbk). xii, 503 p. (2006). ZBL1208.53003.
[3] do Carmo, Manfredo Perdigão, Differential geometry of curves and surfaces, Englewood Cliffs, N. J.: Prentice-Hall, Inc. VIII, 503 p. $ 22.50 (1976). ZBL0326.53001.
[4] do Carmo, Manfredo Perdigão, Riemannian geometry. Translated from the Portuguese by Francis Flaherty, Mathematics: Theory & Applications. Boston, MA etc.: Birkhäuser. xiii, 300 p. (1992). ZBL0752.53001.
