Choose $\varepsilon > 0$ and consider sets defined by
$$\Sigma_k = \{E:\ \left|\int\limits_{E} (f_n-f_m) \right| \leqslant \varepsilon, \textrm{ if } n,m \geqslant k \}$$
Since for any measurable set a limit of integrals exists, we have
$\Sigma = \bigcup\limits_{k} \Sigma_k$. Note, that given an integrable function $f$, the functional $f(E):= \int\limits_{E}f$ is continuous respect to $E$ in metric $\delta$. Indeed, $f(E)-f(F) = \int f \cdot ( 1_E - 1_F )$, hence $|f(E)-f(F)| \leqslant \int |f| \cdot | 1_E - 1_F | = \int |f| \cdot 1_{E \Delta F} = \int\limits_{E\Delta F} |f|$.
The last expression tends to $0$ if $|E\Delta F|$ tends to $0$ because of integrability of $f$, equivalently if $d(E,F)\to 0$. This remark shows that sets $\Sigma_k$ are closed as an intersection of closed subsets of the space $(\Sigma,d)$.
From Baire theorem we obtain that one of sets $\Sigma_k$ has an interior point. Therefore, there exists a measurable set $E_0$ and integer $k$ such that the inequality
$ |f_n(E)-f_m(E) | \leqslant \varepsilon$ holds, whenever $|E\Delta E_0| \leqslant \delta$ and $m,n\geqslant k$. We will show, that this inequality holds in fact for any set $E$, provided that its measure is sufficiency small.
By identities $\mathbf{1}_{E\cup E_0} - \mathbf{1}_{E_0} = \mathbf{1}_{E\cap E_0^{c}}$ and $\mathbf{1}_{E_0}-\mathbf{1}_{E_0\setminus E} = \mathbf{1}_{E\cap E_0}$ we obtain for an arbitrary integrable $f$
$$f(E) = f(E \cap E_0^{c}) + f(E\cap E_0) = f(E\cup E_0) - f(E_0) + f(E_0) - f(E_0\setminus E)$$
If $|E| < \delta$, then all of sets $E_0,E_0\cup E, E_0\setminus E$, belong to the ball $\{E:\ |E\Delta E_0| < \delta \}$. Applying the last inequality to $f_n-f_m$ and $|E| < \delta$ we get
$$ |f_n(E)-f_m(E)| \leqslant 2\varepsilon \quad \textrm { if } |E|<\delta, \ n,m\geqslant k$$
Finally, observe that the finite family of integrable functions $f_1,\ldots, f_k$ is obviously locally uniformly integrable, i.e. $\sup\limits_{i\leqslant k} |f_i(E)| \to 0$ if $|E|\to 0$. Therefore, for sufficiency small $\delta'$ we have
$$|f_i(E)|\leqslant \varepsilon \quad \textrm{ if } |E| < \delta', i\leqslant k$$
Gluing together two estimates that have been derived, we see that for some positive $\delta$
$$ \left|\int\limits_{E} f_n\right| \leqslant 3\varepsilon, \quad \textrm{ if } |E| \leqslant \delta$$
We have estimated integrals for functions $f_n$, instead their modulus. It doesn't matter, however. Namely, applying the last estimate to the set $E\cap \{f_n > 0\}$ and $E\cap \{f_n < 0\}$ respectively (both contained in $E$ hence with a smaller measure), gives finally
$$ \int\limits_{E} |f_n| \leqslant 6\varepsilon, \quad \textrm{ if } |E| \leqslant \delta$$
What finished the proof.
This is an interesting question... which appears to be about "calculus", but which asks for a better answer than could be given in "calculus". And, in fact, I would advocate turning the question around so that the answer to "Can we interchange?" is "Yes, with suitable interpretation...", rather than "Sometimes, but sometimes not."
Upvoted Bob Israel's literal counter-example! :)
An easier analogue is "Clairault's theorem" (I am not at all confident of the correctness of this attribution) about equality of mixed second partials: functions like $xy/(x^2+y^2)$ have unequal mixed second partials at $0$ ... but, from a distributional viewpoint the mixed second partials are absolutely equal. The "discrepancy" is in caring about pointwise values. (Meanwhile, there is also the opposite hazard of thinking that "almost everywhere 0" is operationally $0$ for distributional computations.)
My point is that we oughtn't differentiate unless differentiation is a continuous operation on whatever space of functions, and we oughtn't integrate unless the integration is a continuous map on suitable spaces of functions... so that the question of interchange would be foregone... :)
(A happy situation is when the integrand is a continuous compactly-supported function-valued function of the parameter, and that diff'n is a continuous map from one function-space to another, and then a Gelfand-Pettis integral discussion gives the interchangeability for general reasons. This kind of thing is why L. Schwartz put a perhaps-surprising emphasis on the point that what we now call "Schwartz functions" are functions which admit a smooth extension to a certain $n$-sphere compactification of Euclidean n-space. So, truly, by now, if a situation resists compactification and application of Gelfand-Pettis ideas, perhaps there is a genuine problem. I do also note that "Bochner" or "strong" (which seems to mean "constructed by analogy with Riemann integrals") integrals do not overcome these problems.)
Reprise: there are certainly counter-examples to the literal question, but I'd recommend revising that question exactly in light of the nature of the counter-examples. :)
Best Answer
Yes, we need the boundedness of $|f|^p \ln|f|$ when both $x$ and $p$ vary, but important note is that $p$ may vary on a given segment $[p_1,p_2]$, $1<p_1<p_2$ (this would imply that the derivative $d/dp(\int)$ exists at all points $p\in (p_1,p_2)$, and since the segment is arbitrary, it exists for all $p>1$.) Now this boundedness is pretty clear.