$\DeclareMathOperator\E{E}\DeclareMathOperator\Var{Var}\DeclareMathOperator\P{P}$Note that  
$$\int_{[0,1)^d}\|x\|_2^p\,dx
=\E S_d^{p/2},\tag{1}\label{1}$$
where $S_d:=\sum_1^d U_j^2$ and the $U_j$'s are iid random variables uniformly distributed on the interval $[0,1]$. 

Note next that $\E S_d=d/3$ and $\Var S_d=4d/45<d/10$. So, by [Cantelli's inequality][1],
$$\P(S_d\ge d/6)\ge1-\frac{\Var S_d}{\Var S_d+(d/3-d/6)^2} \\ 
\ge 1-\frac{d/10}{d/10+(1/3-1/6)^2 d} \\
=1-\frac{1/10}{1/10+(1/3-1/6)^2}=:c\in(0,1).$$ 
So, 
$$\E S_d^{p/2}\ge(d/6)^{p/2} \P(S_d\ge d/6)
\ge c(d/6)^{p/2}$$
and hence
$$\Bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\Bigr)^{1/p}
=(\E S_d^{p/2})^{1/p}
\ge c^{1/p}\sqrt{d/6}
\ge c\sqrt{d/6}$$
for $p\ge1$. 

So, the trivial upper bound $\sqrt d$ on $\bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\bigr)^{1/p}$ is optimal up to a universal constant factor.

--- 

For $p\ge2$, one can can do with a much simpler reasoning: by Jensen's inequality, 
$$\Bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\Bigr)^{1/p}
\ge\Bigl(\int_{[0,1)^d}\|x\|_2^2\,dx\Bigr)^{1/2}
=\sqrt{d/3}.$$

---

One may also note that, for any real $p>0$, by the Fatou lemma,
$$\liminf_{d\to\infty}\Bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\Bigr)^{1/p}
\Big/\sqrt{d/3}\ge1.$$
In view of \eqref{1}, this follows because, by the law of large numbers, $S_d/d\to \E U_1^2=1/3$ in probability (as $d\to\infty$).


  [1]: https://en.wikipedia.org/wiki/Cantelli%27s_inequality