ADDENDUM. Vokram told me that I answered another question, not his question. Therefore, I give another counterexample (not so different from the previous one) disproving the equality.
Choose a function $u$ depending only on the second variable, namely $u(x,a) = v(a)$ for all $(x,a) \in X \times A$, so the conditional expectations with regard to $\mathcal{A}$ have no effect on the random variables $U_f : (a,x) \mapsto u(x,f(a)) = v(f(a))$ and $\sum_f U_f$.
We are led to compare $$\sup_{f\in \mathcal{F}} \int_{X\times A} v(f(a)) \; d\pi(x,a) \text{ and } \int_{X\times A}\; \sup_{f\in \mathcal{F}} v(f(a)) \; d\pi(x,a),$$ Calling $\nu$ the second marginal of $\pi$, we compare $$\sup_{f\in \mathcal{F}} \int_{A} v(f(a)) \; d\nu(a) \text{ and } \int_{A} \sup_{f\in \mathcal{F}} v(f(a)) \; d\nu(a).$$
If $A=\mathbb{U}$, unit circle in $\mathbb{C}$, $\nu$ is the Haar measure on $\mathbb{U}$, $\mathcal{F}$ be the family of all rotations on $\mathbb{U}$, and $v : z \mapsto \Re(z)$, then the left-hand side is $0$, whereas the right-and side is $1$ because for all $a \in \mathbb{U}$ $$\sup_{f \in \mathcal F} \Re(f(a)) = |f(a)| = 1.$$
The notations are not good. For example, when you write $$\int_{X\times A}\; \sup_{f\in \mathcal{F}}E \big[u\big(x,f(a)\big) \big| \mathcal{A}\big] d\pi(x,a),$$ $u(x,f(a))$ in the integral behaves like a constant so $E \big[u\big(x,f(a)\big) \big| \mathcal{A}\big] = u(x,f(a))$.
You should write instead $$\int_{X\times A}\; \sup_{f\in \mathcal{F}} E \big[U_f \big| \mathcal{S}\big] d\pi,$$ where $U_f$ the random variable defined bu $U_f(x,a) = u(x,f(a))$ and $\mathcal{S} = \{\emptyset,X\} \times \mathcal{A}$.
Anyway, the equality cannot be always true. For example, consider $X=A=\mathbb{U}$, unit circle in $\mathbb{C}$. Endow $X \times A = \mathbb{U}^2$ with its Borel $\sigma$-field and $\eta \otimes \eta$, where $\eta$ is the Haar measure on $\mathbb{U}$. Then taking conditional expectation with regard to $\mathcal{S}$ is just taking averages over the fist component.
Let $\mathcal{F}$ be the family of all rotations on $\mathbb{U}$, and $u : (x,y) \mapsto \Re(xy)$. Then for each $f \in \mathcal{F}$, the average over $x$ of $U(x,f(a)) = \Re(xf(a))$ is $0$. Yet, since $$\sup_{f \in \mathcal F} u(x,f(a)) = \sup_{y \in \mathbb U} \Re(xy) = |x| = 1,$$ the average over $x$ of $\sup_{f \in \mathcal F} u(x,f(a))$ is $1$.