Here is what I have thought so far about this problem.

Suppose $\Phi$ satisfies the assumptions of de la Vallée-Poussin theorem, I will prove that we can find $\Psi:[0,\infty)\to[0,\infty)$ (satisfying the same assumptions) such that $\{\Psi(u):u\in\mathcal F \}$ is uniformly integrable since this would imply equi-integrability as well.

I claim that, from a given $\Phi$, we can construct a strictly increasing $\Psi:[0,\infty)\to[0,\infty)$ which is convex and superlinear satisfying an extra assumption which is
$$\frac {\Phi(t)}{\Psi(t)}\to \infty \quad\text{as}\quad t\to\infty.$$

Assuming that the claim is true, for any $\varepsilon>0$ we can find $T$ such that $\Psi(t)\le \varepsilon \Phi(t)$ whenever $t>T$. Take $M:=\Psi(T)$ and note that since $\Psi$ is strictly increasing, $\Psi(t)>M \iff t>T$. Then
$$\begin{align}
\sup_{u\in\mathcal F} \int_{\{ \Psi(|u|) >M \}} \Psi(|u(x)|)\,dx
&= \sup_{u\in\mathcal F} \int_{\{ |u|>T \}} \Psi(|u(x)|)\,dx \\
&\le \varepsilon \sup_{u\in\mathcal F} \int_{\{ |u|>T \}} \Phi(|u(x)|)\,dx \\
&\le \varepsilon \sup_{u\in\mathcal F} \int_{\Omega} \Phi(|u(x)|)\,dx \\
\end{align}$$
which proves the uniform integrability of $\{\Psi(u):u\in\mathcal F \}$.

Proving the claim is a bit longer, but essentially I constructed a piecewise-linear convex function $\Psi$ by interpolating some chosen points of $\log(\Phi)$ and some extra modification to make it convex.

If you see any error in my reasoning, please mention it so that I can attempt to fix it. Thank you very much.