Okay, I think I've worked out that the answer is no, i.e. there exists a sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{B}$ such that $\mathbb{E}_\mathcal{G}$ is not universally measurable.

(We will write $\lambda$ for the Lebesgue measure on $([0,1],\mathcal{B})$.)

It is well-known that the evaluation $\sigma$-algebra is precisely the Borel $\sigma$-algebra of the topology of weak convergence, which is a Polish topology. So then, to show that (for some given $\mathcal{G}$) $\mathbb{E}_\mathcal{G}$ is not universally measurable, it is sufficient (by Lusin's theorem) to find a probability measure $Q$ on $\mathcal{M}_2$ such that for every measurable set $E \subset \mathcal{M}_2$ with $Q(E)>0$, the restriction of $\mathbb{E}_\mathcal{G}$ to $E$ is not a continuous function (with respect to the topology of weak convergence).

Let $A \subset [0,1]$ be a set such that $A$ and $[0,1] \setminus A$ intersect every Lebesgue-positive measure set. (Assuming the axiom of choice, such a set $A$ exists, as shown here.) Let $\mathcal{G}$ be the $\sigma$-algebra consisting of all countable subsets of $A$ and their complements.

Let $Q$ be the image measure of $\lambda \otimes \lambda$ under the map $D : (x,y) \mapsto \frac{1}{2}(\delta_{x,x}+\delta_{y,y})$. Let $E \subset \mathcal{M}_2$ be any measurable set with $Q(E)>0$, and let $F=D^{-1}(E)$. (So $\lambda \otimes \lambda(F)>0$.) Let $F'=\{x \in [0,1] : \lambda(y:(x,y)\in F)>0\}$. Obviously $\lambda(F')>0$, so fix a point $x \in F' \setminus A$. For any $y \in [0,1]$, we have that $$ \mathbb{E}_\mathcal{G}(D(x,y)) \ = \ \left\{ \begin{array}{c l} \frac{1}{2}(\delta_{x,x}+\delta_{y,y}) & y \in A \\ \frac{1}{4}(\delta_{x,x} + \delta_{x,y} + \delta_{y,x} + \delta_{y,y}) & y \not\in A. \end{array} \right. $$ So since $Y:=\{y \in [0,1] : (x,y) \in F\}$ has positive Lebesgue measure, it is clear that $\mathbb{E}_\mathcal{G}(D(x,\cdot))$ is not continuous on $Y$, so $\mathbb{E}_\mathcal{G} \circ D$ is not continuous on $F$, so (since $D$ is obviously continuous) $\mathbb{E}_\mathcal{G}$ is not continuous on $E$.

Okay, I think I've worked out that the answer is no, i.e. there exists a sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{B}$ such that $\mathbb{E}_\mathcal{G}$ is not universally measurable.

(We will write $\lambda$ for the Lebesgue measure on $([0,1],\mathcal{B})$.)

It is well-known that the evaluation $\sigma$-algebra is precisely the Borel $\sigma$-algebra of the topology of weak convergence, which is a Polish topology. So then, to show that (for some given $\mathcal{G}$) $\mathbb{E}_\mathcal{G}$ is not universally measurable, it is sufficient (by Lusin's theorem) to find a probability measure $Q$ on $\mathcal{M}_2$ such that for every measurable set $E \subset \mathcal{M}_2$ with $Q(E)>0$, the restriction of $\mathbb{E}_\mathcal{G}$ to $E$ is not a continuous function (with respect to the topology of weak convergence).

Let $A \subset [0,1]$ be a set such that $A$ and $[0,1] \setminus A$ intersect every Lebesgue-positive measure set. (Assuming the axiom of choice, such a set $A$ exists, as shown here.) Let $\mathcal{G}$ be the $\sigma$-algebra consisting of all countable subsets of $A$ and their complements.

Let $Q$ be the image measure of $\lambda \otimes \lambda$ under the map $D : (x,y) \mapsto \frac{1}{2}(\delta_{x,x}+\delta_{y,y})$. Let $E \subset \mathcal{M}_2$ be any measurable set with $Q(E)>0$, and let $F=D^{-1}(E)$. (So $\lambda \otimes \lambda(F)>0$.) Let $F'=\{x \in [0,1] : \lambda(y:(x,y)\in F)>0\}$. Obviously $\lambda(F')>0$, so fix a point $x \in F' \setminus A$. For any $y \in [0,1]$, we have that $$ \mathbb{E}_\mathcal{G}(D(x,y)) \ = \ \left\{ \begin{array}{c l} \frac{1}{2}(\delta_{x,x}+\delta_{y,y}) & y \in A \\ \frac{1}{4}(\delta_{x,x} + \delta_{x,y} + \delta_{y,x} + \delta_{y,y}) & y \not\in A. \end{array} \right. $$ So since $Y:=\{y \in [0,1] : (x,y) \in F\}$ has positive Lebesgue measure, it is clear that $\mathbb{E}_\mathcal{G}(D(x,\cdot))$ is not continuous on $Y$, so $\mathbb{E}_\mathcal{G} \circ D$ is not continuous on $F$, so (since $D$ is obviously continuous) $\mathbb{E}_\mathcal{G}$ is not continuous on $E$.