The measure-theoretic formulation of conditional probability/expectation is cumbersome, so it is often easy to lose connection between intuitive facts and the formal demonstrations of them. I personally often have difficulty with arguing for seemingly obvious facts.

In this case, I think more than your theorem is true: your function $EX(\omega)$ is a regular version of the conditional expectation of $f(\omega):=X(\omega,\omega)$ given $\mathcal{G}$, i.e., $\mathbb{E}[f\,|\,\mathcal{G}]$.  By definition, this means that $EX(\omega)$ is $\mathcal{G}$-measurable and
\begin{align}
   \underbrace{\int_G EX(\omega)\,\mathrm{d}\mathbb{P}(\omega)}_{\int_G \mathbb{E}[f\,|\,\mathcal{G}]\,\mathrm{d}\mathbb{P}(\omega)} &=
   \underbrace{\int_G X(\omega,\omega)\, \mathrm{d}\mathbb{P}(\omega)}_{\int_G f(\omega) \mathrm{d}\mathbb{P}(\omega)} \;,
\end{align}
for every $G\in\mathcal{G}$.

This is almost immediate from your definition. Note that

$\quad$($\star$) for each $\omega$, we have $X(\omega,v)=X(v,v)$ for $\mu_\omega$-almost every $v$.

Thus,
\begin{align}
   \int_G EX(\omega)\,\mathrm{d}\mathbb{P}(\omega) &=
      \int_G\int f(v)\,\mathrm{d}\mu_\omega(v)\mathrm{d}\mathbb{P}(\omega) \\
   &= \int_G f(v)\,\mathrm{d}\mathbb{P}(\omega) \;.
\end{align}