It's not true. For a counterexample, take $\Omega = \{a,b,c\}$ to be a sample space with 3 points, $\mathcal{F} = 2^{\Omega}$, and $P(A) = \frac{1}{3} |A|$ to be the uniform probability measure assigning probability 1/3 to each outcome. Let's represent a random variable $X : \Omega \to \mathbb{R}$ as the ordered triple $(X(a), X(b), X(c))$.
Set $\mathcal{G}_1 = \{\Omega, \emptyset, \{a\}, \{b,c\}\}$, and $\mathcal{G}_2 = \{\Omega, \emptyset, \{b\}, \{a,c\}\}$. Let $X$ be the random variable $(1,2,3)$. Then one can directly compute $$\begin{align*} E[X \mid \mathcal{G}_1] &= (1, 2.5, 2.5) \\ E[X \mid \mathcal{G}_2] &= (2,2,2) \\ E[E[X \mid \mathcal{G}_1] \mid \mathcal{G}_2] &= (1.75,2.5,1.75) \\ E[E[X \mid \mathcal{G}_2] \mid \mathcal{G}_1] &= (2,2,2). \end{align*}$$