Note that  
$$\int_{[0,1)^d}\|x\|_2^p\,dx
=ES_d^{p/2},\tag{1}$$
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 $ES_d=d/3$ 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, 
$$ES_d^{p/2}\ge(d/6)^{p/2} P(S_d\ge d/6)
\ge c(d/6)^{p/2}$$
and hence
$$\Big(\int_{[0,1)^d}\|x\|_2^p\,dx\Big)^{1/p}
=(ES_d^{p/2})^{1/p}
\ge c^{1/p}\sqrt{d/6}
\ge c\sqrt{d/6}$$
for $p\ge1$. 

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

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

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One may also note that, for any real $p>0$, by the Fatou lemma,
$$\liminf_{d\to\infty}\Big(\int_{[0,1)^d}\|x\|_2^p\,dx\Big)^{1/p}
\Big/\sqrt{d/3}\ge1.$$
In view of (1), this follows because, by the law of large numbers, $S_d/d\to EU_1^2=1/3$ in probability (as $d\to\infty$).


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