The answer is yes, thanks to Kakashi's idea in the comments above of looking at convergence in measure.
Indeed: for $\varepsilon > 0$, consider $E_{n,\varepsilon} := \{(x,t) \in \Omega \times [0,T] \mid |f(x,t) - f_n(x,t)| \geq \varepsilon\}$, and denote by $\lambda$ all the Lebesgue measures.
First, consider the case $q \geq p$.
Then, we get:
$$\begin{split}\varepsilon^p \lambda(E_{n,\varepsilon}) &= \int_{E_{n,\varepsilon}} \varepsilon^p \,\mathrm{d}t\mathrm{d}x\\
&\leq \int_{E_{n,\varepsilon}} |f(x,t) - f_n(x,t)|^p \,\mathrm{d}t\mathrm{d}x\\
&\leq \iint_{\Omega \times [0,T]} |f(x,t) - f_n(x,t)|^p \,\mathrm{d}x\mathrm{d}t\end{split}$$
Fubini-Tonelli lets us change the order of integration, and we can apply $x \mapsto x^{\frac{q}{p}}$ to the inequality, hence:
$$\left(\varepsilon^p \lambda(E_{n,\varepsilon})\right)^{\frac{q}{p}} \leq \left(\int_{[0,T]} \left(\int_{\Omega}|f(x,t) - f_n(x,t)|^p \,\mathrm{d}x\right)\mathrm{d}t\right)^{\frac{q}{p}}$$
Now we can apply Jensen's inequality on $[0,T]$ of finite measure $T$ with the function $x \mapsto x^\frac{q}{p}$, which is convex due to the assumption that $q \geq p$, which provides:
$$\begin{split} 0 \leq \left(\varepsilon^p \lambda(E_{n,\varepsilon})\right)^{\frac{q}{p}} &\leq \left(\int_{[0,T]} \left(\int_{\Omega}|f(x,t) - f_n(x,t)|^p \,\mathrm{d}x\right)\mathrm{d}t\right)^{\frac{q}{p}}\\
&\leq T^{\frac{q}{p}}\left(\frac{1}{T}\int_{[0,T]} \left(\int_{\Omega}|f(x,t) - f_n(x,t)|^p \,\mathrm{d}x\right)\mathrm{d}t\right)^{\frac{q}{p}}\\
&\leq T^{\frac{q}{p} - 1}\int_{[0,T]} \left(\int_{\Omega}|f(x,t) - f_n(x,t)|^p \,\mathrm{d}x\right)^{\frac{q}{p}}\mathrm{d}t\\
&\leq T^{\frac{q}{p} - 1} \|f - f_n\|^q_{L^q(L^p)} \xrightarrow[n \to \infty]{} 0\end{split}$$
This gives us $\lambda(E_{n,\varepsilon}) \to 0$ when $n \to \infty$ for all $\varepsilon > 0$, thus $(f_n)_n$ converges globally in measure to $f$, which means that $(f_n)_n$ admits a subsequence which converges almost everywhere to $f$.
Now, consider the case $q < p$.
Similarly to what was done before, we'll still use Jensen's inequality, but this time "the other way round" with the concave function $x \mapsto x^{\frac{q}{p}}$. However, since that'll be done on the $\Omega$-integral so to speak, we'll need this time to consider local convergence in measure.
Pick $F \subset \Omega \times [0,T]$ of finite measure, and $F_1 \subset \Omega$ of finite measure such that $F \subset F_1 \times [0,T]$. Finally, instead of $E_{n, \varepsilon}$, define $E_{n, \varepsilon, F} := \{(x,t) \in F \mid |f(x,t) - f_n(x,t)| \geq \varepsilon\}$.
Then, we'll have:
$$\begin{split}\varepsilon^q \lambda(E_{n,\varepsilon, F}) &= \int_{E_{n,\varepsilon, F}} \varepsilon^q\,\mathrm{d}t\mathrm{d}x\\
&\leq \int_{E_{n,\varepsilon, F}} |f(x,t) - f_n(x,t)|^q \,\mathrm{d}t\mathrm{d}x\\
&\leq \iint_{F_1 \times [0,T]} |f(x,t) - f_n(x,t)|^q \,\mathrm{d}x\mathrm{d}t\\
&\leq \lambda(F_1)\int_{[0,T]} \frac{1}{\lambda(F_1)} \left(\int_{F_1} |f(x,t) - f_n(x,t)|^{p\frac{q}{p}}\,\mathrm{d}x\right)\mathrm{d}t\\
&\overset{\text{Jensen}}{\leq} \lambda(F_1) \int_{[0,T]} \left(\frac{1}{\lambda(F_1)}\int_{F_1} |f(x,t) - f_n(x,t)|^p \,\mathrm{d}x\right)^{\frac{q}{p}}\mathrm{d}t\\
&\leq \lambda(F_1)^{1 - \frac{q}{p}}\|f_n - f\|^q_{L^q(L^p)}\end{split}$$
Thus we conclude in a very similar fashion as for the previous case, as local convergence in measure still grants the existence of the desired subsequence when the measure is $\sigma$-finite, which is the case here.
Best Answer
We recall a few facts:
We say $f_n \xrightarrow[]{\text{measure}} f$ if $$\lambda(|f_n - f| > \epsilon) \xrightarrow[]{n \rightarrow \infty}0,$$ where $\lambda$ is Lebesgue measure.
Convergence in measure implies convergence almost everywhere of a subsequence
The goal is to show that $f_n \xrightarrow[]{\|\cdot\|_2} 0$ implies that $f_n \xrightarrow[]{\text{measure}}0$, where $$\|f\|_2 := \left( \int_\mathbb{R} |f|^2 d\lambda \right)^{1/2}.$$ The result follows immediately once we have this.
Notice that $f_n \xrightarrow[]{\|\cdot\|_2} 0$ implies that $$\lim_{n \rightarrow \infty} \int_{\mathbb{R}} |f_n|^2 d\lambda = 0.$$ Let $A(n,\epsilon) := \{x \in \mathbb{R} \ | \ |f_n(x)| > \epsilon\}$. Then we have
$$ \int_{\mathbb{R}} |f_n|^2 d\lambda \geq \int_{A(n,\epsilon)} |f_n|^2 d\lambda > \int_{A(n,\epsilon)} \epsilon^2 d\lambda = \epsilon^2 \lambda(A(n,\epsilon)) = \epsilon^2 \lambda( |f_n| > \epsilon),$$
so
$$\lambda(|f_n| > \epsilon) < \frac{\|f_n\|_2^2}{\epsilon^2}.$$
Now take $n \rightarrow \infty$ and we have convergence in measure.