No. For a very simple example, take $\Omega = \{a,b,c\}$ consisting of three points, with $\mathcal{F} = 2^\Omega$ and $P(A) = |A|/3$ the uniform measure. Let $\mathcal{G} = \{\{a\}, \{b,c\}, \Omega, \emptyset\}$. Then $\mathcal{G}$ is a proper sub-$\sigma$-algebra but no nontrivial event in $\mathcal{F}$ is independent of it.
Edit: Andres Caicedo asks for a non-atomic example. George Lowther gave one. Another is supplied by taking $\Omega = [0,1]$, $\mathcal{G} = \mathcal{B}_{[0,1]}$ the Borel $\sigma$-field, $\mathcal{F} = \mathcal{L}$ the Lebesgue $\sigma$-field which is the completion of $\mathcal{G}$, and $P = $ Lebesgue measure. Now by definition of $\mathcal{L}$, for any $A \in \mathcal{F}$ we have $A = B \cup N$ where $B \in \mathcal{G}$ and $P(N) = 0$. Then $P(A \cap B) = P(B) = P(A)$ so $A$ and $B$ are independent iff $P(A) = 0$ or $1$.
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 $E$.
Best Answer
The answer to the first question is yes. There is a class of probability spaces known under various names such as superatomless, saturated, nowhere countably-generated, $\aleph_1$-atomless, and a couple of other names that have exactly this property. Note that the restriction to sub-$\sigma$-algebras admitting sets of interior measure is superfluous. Otherwise, the $\sigma$-algebra $\Sigma$ would be trivially independent of it. This paper might be a good entry to the topic.
A typical example of such a probability space would be the independent product measure on uncountably many copies of the unit interval endowed with the uniform distribution. It is the canonical example in a special sense. If you take a probability space $(\Omega,\Sigma,\pi)$ and identify two measurable sets $A$ and $B$ with $\pi(A\Delta B)=0$, you obtain the so-called measure algebra. There is a fundamental theorem due to Dorothy Maharam that for every probability space, the measure algebra is a countable weighted sum of product measures obtained from $[0,1].$ I think one might also be able to use this to prove that the second part of the question holds true.