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}

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 $v$ (see @1 below).

Thus, $EX(\omega)=\int f(v)\,\mathrm{d}\mu_\omega(v)$. That $\int f(v)\,\mathrm{d}\mu_\omega(v)$ is $\mathcal{G}$-measurable for every $f$ is standard (see @2 below). Furthermore, \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}(v) \;. \end{align}

Further details:

@1: I believe this can be shown using the usual measure theory trick. Namely, first show that ($\star$) holds if $X$ is the indicator of a rectangle $G\times F$ with $G\in\mathcal{G}$ and $F\in\mathcal{F}$. Then, show that the family of sets $A\in\mathcal{G}\otimes\mathcal{F}$ for which $X(\cdot,\cdot):=1_A$ satisfies ($\star$) is a $\sigma$-algebra. Then, show that $(\star)$ holds if $X$ is a simple function. Finally, show that if $X$ is the uniform limit of an increasing sequence of functions satisfying ($\star$), then $X$ itself satisfies ($\star$).

@2: The argument for this again uses the usual trick. We already know (by definition) that $\int f(v)\,\mathrm{d}\mu_\omega(v)$ is $\mathcal{G}$-measurable if $f$ is an indicator function. First, show that the same holds for simple function. Then, show that if $f$ is the uniform limit of an increasing sequence $f_1\leq f_2\leq\cdots$ such that $\int f_1(v)\,\mathrm{d}\mu_\omega(v)$ is $\mathcal{G}$-measurable, then $\int f(v)\,\mathrm{d}\mu_\omega(v)$ is also $\mathcal{G}$-measurable.