Suppose I parametrize the 2-sphere as
\begin{align} \vec{r}(u,v)= \begin{bmatrix} \cos u\sin v\\ \sin u\sin v\\ \cos v \end{bmatrix}\ ,\ (u,v)\in[0,2\pi]\times[0,\pi] \end{align}
and I have a curve lying on the sphere
\begin{align} \vec{\gamma}(t)= \begin{bmatrix} \cos u(t)\sin v(t)\\ \sin u(t)\sin v(t)\\ \cos v(t) \end{bmatrix}\ ,\ t\in[0,2\pi]. \end{align}
I want to find the mean distance from a point on the curve to a point on the sphere, i.e., the following integral
\begin{align} \frac{1}{4\pi}\iint\limits_{S^2}\arccos\bigl(\vec{\gamma}(t)\cdot\mathbf{x}\bigr)\ d\mathbf{x}&=\frac{1}{4\pi}\iint\limits_{[0,2\pi]\times[0,\pi]}\arccos\bigl(\vec{\gamma}(t)\cdot\vec{r}(u,v)\bigr)\ \left\|\frac{\partial \vec{r}}{\partial u}\times\frac{\partial \vec{r}}{\partial v}\right\|dudv\\\\ &=\frac{1}{4\pi}\int\limits_0^\pi\int\limits_0^{2\pi}\arccos\left(\begin{bmatrix} \cos u(t)\sin v(t)\\ \sin u(t)\sin v(t)\\ \cos v(t) \end{bmatrix}\cdot\begin{bmatrix} \cos u\sin v\\ \sin u\sin v\\ \cos v \end{bmatrix}\right)\sin v\ dudv. \end{align}
If $t$ is fixed, I understand how to carry out the integral (at least in theory because it probably won't have a closed form). My question is, do I need to add another integral sign in front to calculate the mean distance I'm looking for, like so:
\begin{align} \frac{1}{4\pi}\int\limits_0^{2\pi}\int\limits_0^\pi\int\limits_0^{2\pi}\arccos\left(\begin{bmatrix} \cos u(t)\sin v(t)\\ \sin u(t)\sin v(t)\\ \cos v(t) \end{bmatrix}\cdot\begin{bmatrix} \cos u\sin v\\ \sin u\sin v\\ \cos v \end{bmatrix}\right)\sin v\ dudvdt. \end{align}
This seems like what I ought to do but I'd like some corroboration.
The average (mean) value of a continuous/integrable function $f$ on $[a,b]$ is, by definition, $$\frac1{b-a} \int_a^b f(t)\,dt.$$ So, yes, you need the third integral and another factor of $\frac1{2\pi}$. (I'm not quite sure why you're parametrizing your curve specifically by $[0,2\pi]$. Wouldn't $[0,1]$ do just as well?)