Suppose that $(f_m)_m, f \subset L^p(\Omega)$ are nonnegative, such that $f_m \rightarrow f$ in $L^p$; that is:
$$ \|f_m - f\|_p \rightarrow 0 $$
It is clear that $({f_m}^r)_m, f^r \subset L^{\frac{p}{r}}(\Omega)$, since $\int |f|^p = \int|f^r|^{\frac{p}{r}}$. Can we conclude that ${f_m}^r \rightarrow f^r$ in $L^{\frac{p}{r}}$? That is:
$$ \left\| {f_m}^r - f^r \right\|_{\frac{p}{r}} \rightarrow 0 $$
Note: we assume $1<r<p<\infty$, and $\Omega \subset \mathbb{R}^n$ is measurable.
We can use the following result: if a sequence $\left(g_n\right)_{n\geqslant 1}$ is such that $g_n\to g$ almost everywhere and $\int\left\lvert g_n\right\rvert^q\to \int\left\lvert g\right\rvert^q$ then $\int\left\lvert g_n-g\right\rvert^q\to 0$. This follows from an application of Fatou's lemma to the sequence $\left(h_n\right)_{n\geqslant 1}$ defined by $h_n=2^{q-1}\left(\left\lvert g_n\right\rvert^q +\left\lvert g\right\rvert^q\right)-\left\lvert g_n-g\right\rvert^q$.
Extract an almost everywhere convergent subsequence $\left(f_{n_k}\right)$ of $\left(f_n\right)$, $q=p/r$, $g_k:=\left\lvert f_{n_k}\right\rvert^r$ to get that $\left\| {f_{n_k}}^r - f^r \right\|_{\frac{p}{r}} \rightarrow 0$. Apply this reasoning to an arbitrary subsequence of $\left(f_n\right)$ instead of of $\left(f_n\right)$ to get the wanted result. This shows that for all subsequence, you can extract a further subsequence which converges to $f^r$ in $L^{p/r}$, which gives the wanted result (note that in the link, convergence of sequences is treated but it works with the same proof for normed spaces).
The answer is yes. The proof of this fact relies on the following lemma:
Let $E \subset \mathbb{R}^n$ be measurable, and $1<p<\infty$. Then the function $\phi_p(u) = |u|^{p-1} u$ defines a homeomorphism from $L^p(E)$ to $L^1(E)$. Moreover, $\forall u,v \in L^p(E)$:
$$ 2^{1-p} \|u-v\|_p^p \leq \|\phi_p(u) - \phi_p(v)\|_1 \leq p (\|u\|_p+\|v\|_p)^{p-1} \|u-v\|_p $$
This lemma follows from exercises 46-48 of Royden & Fitzpatrick's Real Analysis (4th edition), p.154 (end of chapter 7).
The result follows easily from here. Since the ${f_m}^r$ and $f^r$ are in $L^{\frac{p}{r}}$, we use the left inequality with the value $p/r$ to get: $$ 2^{1-\frac{p}{r}} {\|{f_m}^r - f^r \|_{\frac{p}{r}}}^{\frac{p}{r}} \leq \| \phi_{\frac{p}{r}}({f_m}^r) - \phi_{\frac{p}{r}}(f^r) \|_1 = \| {f_m}^p - f^p \|_1 $$ Then we use the right inequality with the value $p$ to get: $$ \|{f_m}^p - f^p\|_1 = \| \phi_p(f_m) - \phi_p(f) \|_1 \leq p (\|f_m\|_p+\|f\|_p)^{p-1} \|f_m-f\|_p $$
Finally, since $f_m \rightarrow f$ in $L^p$, we can bound $\|f_m\|_p$ uniformly. Putting this together with the inequalities above, we conclude that there are constants $C_1, C_2 > 0$ such that: $$ C_1 {\|{f_m}^r - f^r \|_{\frac{p}{r}}}^{\frac{p}{r}} \leq C_2 \|f_m-f\|_p $$ So, as $m \rightarrow \infty$, we know by assumption that the right-hand side goes to $0$, and it is clear that this implies that the left-hand side goes to $0$. Hence ${f_m}^r \rightarrow f^r$ in $L^{\frac{p}{r}}$.