I have asked this question some hours ago but I have noticed that I have not provided enough details and didnt receive any answers. I think I have now found a solution to the problem and I hope someone tell me if I am wrong somewhere or not.
A function $f:[a,b]\rightarrow \mathbb{R},a<b\in\mathbb{R}$ is integrable iff for every $\epsilon$ there exists $\psi,\phi\in \mathcal{T}[a,b]$ stepfunctions such that $\phi(x)\leq f(x) \leq \psi(x),\forall x\in[a,b]$ (from now on I am just gonna write $\phi\leq f \leq \psi$ for this) and $\int_{a}^{b}(\psi-\phi)(x)dx\leq \epsilon$ (You can assume that the definition of an integral of a stepfunction is already defined and that the sum i.e. product with a scalar is again a stepfunction).
From this equivalance we can deduce that if $f$ is integrable then $f$ is necessarily bounded.
We also already know that if $f$ is integrable so is $|f|$. (Note all the functions I am talking about are from now on always defined on some interval $[a,b]$)
The proof for the statement in the title:
Since $|f|$ is integrable we can assume there exists a $C\in\mathbb{R_+}$ such that $|f|\leq C$ then because the function is integrable there also exists $\phi,\psi$ such that $\phi\leq |f| \leq \psi$ and $\int_{a}^{b}(\psi-\phi)(x)dx\leq \frac{\epsilon}{pC^{p-1}}$. We can additionaly assume that $0< \phi\leq |f| \leq \psi < C$. We now look at the function $x^{p}:[0,C]\rightarrow \mathbb{R}$ then $\frac{d}{dx}x^p=x^{p-1}p\leq C^{p-1}p$. By meanvalue theorem we then can say $\psi^{p}-\phi^{p}\leq (\psi-\phi)pC^{p-1}$.
This finaly gives us $\int_{a}^{b}(\psi^p-\phi^p)(x)dx\leq pC^{p-1}\int_{a}^{b}(\psi-\phi)(x)dx\leq\epsilon$.
Therefore with $\psi^p,\phi^p$ we have provided two fitting functions. I would really appreciate if someone could tell me if this is right or wrong.
In case you are interested in a more general result: If $f:[a,b]\to [c,d]$ is Riemann integrable, and $g:[c,d]\to \mathbb R$ is continuous, then $g\circ f$ is Riemann integrable on $[a,b].$ In your problem we can take $g(x)=|x|^p$ to get the desired result. (And this would be true for all nonnegative $p,$ not just $p\ge 1.$)