It is a purely measure-theoretic result that this holds for almost all initial points. As I describe below, I think this implies that it actually holds for all initial points, in the following sense. Call $\mathbb P_{x_0}$ the distribution of the random walk started at $x_0$, and $\mu$ the uniform measure on $S$.
For all $x_0\in S$, for all $f_0:S\to\mathbb R$ continuous, we have $$ \lim_{n\to\infty}\frac1n\sum_{i<n}f_0(x_i) = \int fd\mu $$ $\mathbb P_{x_0}$-almost surely.
The fact that $\delta$ is small enough plays no role in the first part.
Let $\Omega\subset S^{\mathbb N}$ be the set of sequences such that two consecutive terms are at distance precisely $\delta$. This is a compact space (closed subset of a compact space, by Tychonoff or simply using a convenient metric), and it carries a natural probability $\mathbb P_\mu$, corresponding to $x_0$ being distributed according to $\mu$ and the following steps according to the random walk rules. The shift operator $$ T:(x_0,x_1,\ldots)\mapsto(x_1,x_2,\ldots) $$ is such that $T_*\mu=\mu$ (let us just accept this until the end of the proof sketch), so $(\Omega,\mathbb P_\mu,T)$ is a dynamical system in the measure-theoretic sense. According to Birkhoff's ergodic theorem, for $\mathbb P_\mu$-almost every $x$, we have $$ \lim_{n\to\infty}\frac1n\sum_{i<n}f(T^ix) = \mathbb E[f|\mathcal F_T](x) $$ for all $f:\Omega\to\mathbb R$ continuous, where $\mathcal F_T$ is the algebra of $T$-invariant sets. Now if $(\Omega,\mathbb P_\mu,T)$ is ergodic, then $$\mathbb E[f|\mathcal F_T](x) = \int fd\mathbb P_\mu = \int f_0d\mu $$ for all $f$ depending only on $x_0$, i.e. $f:x\mapsto f_0(x)$. Thus my claim will follow by Fubini.
It remains to prove that $\mu$ is $T$ invariant, and that the resulting system is ergodic. According to the probabilistic interpretation in terms of Markov chains, it suffices to show that $x_1$ has the same distribution as $x_0$ (the uniform one) under $\mathbb P_\mu$. This is a consequence of the fact that the geodesic flow leaves the measure induced on $TM$ invariant, which itself is a consequence of Liouville's theorem because the geodesic flow is Hamiltonian.
Now let us show the system is ergodic. Let $f:\Omega\to\mathbb R$ be $T$-invariant, i.e. $f\circ T = f$, and suppose also that $f$ depends only on the first $k$ terms of the sequence. Let us show that it is constant by choosing $x$ and $y$ arbitrary and showing $f(x)=f(y)$. Let $N$ be a number large enough that $N\delta$ is larger than the diameter of the manifold. Killing the first $k+N$ terms of $x$ using $T$, then adding $N$ terms linking $x_k$ to $y_{k-1}$, and adding the first $k$ terms of $y$, we see that $f(x) = f(y_k)$, where $y_k$ is equal to $y$ up to the $k$th term. Because $f$ only depends on the first $k$ terms, we have in fact $f(x)=f(y)$. Now an approximation argument shows that the system is ergodic (see for instance a previous answer of mine here).
Now this extends I think to all points in the manifold, provided $\delta$ is less than the injectivity radius. Let $A$ be the set of points $x_0$ such that the random walk started at $x_0$ is evenly distributed almost surely, and $\mathbf 1_A$ its indicator function. Then $$ \mathbb P_{x_0}(\text{even distribution}) = \int \mathbb P_{x_2}(\text{even distribution})d\mathbb P_{x_0} = \int 1d T^2_*\delta_{x_0} = 1, $$ because the distribution of $x_2$ under $\mathbb P_{x_0}$ is continuous with respect to the Lebesgue measure (image of the Lebesgue measure on the product of two sphere under a submersion almost everywhere). In fact the result is true for any closed Riemannian manifold, as maybe we could have expected, although we have to look for $x_d$, where $d$ is the dimension of the manifold. The “I think” I use comes from the fact that although $x_d$ is the image of a map from a product of $d$ spheres to $S$, I cannot give a simple argument of the fact (natural to me) that this is a submersion at all points that form a basis of $\mathbb R^d$.