The answer is in the negative, but the problem can be fixed up.
To see what goes wrong, let $X$ have a Student t distribution with two degrees of freedom. Its salient properties are that $\mathbb{E}(|X|)$ is finite but $\mathbb{E}(|X|^2)=\infty$. Consider the bivariate distribution of $(X,X)$. Let $f(x,y)dxdy$ be its distribution element (which is singular: it is supported only on the diagonal $x=y$). Along the diagonal, $||(x,y)||=|x|\sqrt{2}$, whence
$$\mathbb{E}\left(||(X,X)||^1\right) = \mathbb{E}\left(\sqrt{2}|X|\right) \lt \infty$$
whereas
$$\iint x^1 y^1 f(x,y) dx dy = \int x^2 f(x,x) dx = \infty.$$
Analogous computations in $p$ dimensions should make it clear that $$\int\cdots\int |x_1|^k|x_2|^k\cdots |x_p|^k f(x_1,\ldots, x_p)dx_1\cdots dx_p$$
really is a moment of order $pk$, not $k$. For more about multivariate moments, please see Let $\mathbf{Y}$ be a random vector. Are $k$th moments of $\mathbf{Y}$ considered?.
To find out what the relationships ought to be between the multivariate moments and the moments of the norm, we will need two inequalities. Let $x=(x_1, \ldots, x_p)$ be any $p$-dimensional vector and let $k_1, k_2, \ldots, k_p$ be positive numbers. Write $k=k_1+k_2+\cdots k_p$ for their sum (implying $k_i/k \le 1$ for all $i$). Let $q \gt 0$ be any positive number (in the application, $q=2$ for the Euclidean norm, but it turns out there's nothing special about the value $2$). As is customary, write
$$||x||_q = \left(\sum_i |x_i|^q\right)^{1/q}.$$
First, let's apply the AM-GM inequality to the non-negative numbers $|x_i|^q$ with weights $k_i$. This asserts that the weighted geometric mean cannot exceed the weighted arithmetic mean:
$$\left(\prod_i (|x_i|^q)^{k_i}\right)^{1/k} \le \frac{1}{k}\sum_i k_i|x_i|^q.$$
Overestimate the right hand side by replacing each $k_i/k$ by $1$ and take the $k/q$ power of both sides:
$$\prod_i |x_i|^{k_i} = \left(\left(\prod_i (|x_i|^q)^{k_i}\right)^{1/k}\right)^{k/q} \le \left(\sum_i |x_i|^q\right)^{k/q} = ||x||_q^k.\tag{1}$$
Now let's overestimate $||x||_q$ by replacing each term $|x_i|^q$ by the largest among them, $\max(|x_i|^q) = \max(|x_i|)^q$:
$$||x||_q \le \left(\sum_i \max(|x_i|^q)\right)^{1/q} = \left(p \max(|x_i|)^q\right)^{1/q} = p^{1/q} \max(|x_i|).$$
Taking $k^\text{th}$ powers yields
$$||x||_q^k \le p^{k/q} \max(|x_i|^k) \le p^{k/q} \sum_i |x_i|^k.\tag{2}$$
As a matter of notation, write
$$\mu(k_1,k_2,\ldots,k_p) = \int\cdots \int |x_1|^{k_1}|x_2|^{k_2}\cdots|x_p|^{k_p} f(x)\,dx.$$
This is the moment of order $(k_1,k_2,\ldots,k_p)$ (and total order $k$). By integrating aginst $f$, inequality $(1)$ establishes
$$\mu(k_1,\ldots,k_p) \le \int\cdots\int ||x||_q^k f(x)\,dx = \mathbb{E}(||X||_q^{k})\tag{3}$$
and inequality $(2)$ gives $$\mathbb{E}(||X||_q^{k})\le p^{k/q}\left(\mu(k,0,\ldots,0) + \mu(0,k,0,\ldots,0) + \cdots + \mu(0,\ldots,0,k)\right).\tag{4}$$
Its right hand side is, up to a constant multiple, the sum of the univariate $k^\text{th}$ moments. Together, $(3)$ and $(4)$ show
Finiteness of all univariate $k^\text{th}$ moments implies finiteness of $\mathbb{E}(||X||_q^{k})$.
Finiteness of $\mathbb{E}(||X||_q^{k})$ implies finiteness of all $\mu(k_1,\ldots,k_p)$ for which $k_1+\cdots +k_p=k$.
Indeed, these two conclusions combine as a syllogism to show that finiteness of the univariate moments of order $k$ implies finiteness of all multivariate moments of total order $k$.
Thus,
For all $q \gt 0$, the $k^\text{th}$ moment of the $L_q$ norm $\mathbb{E}(||X||_q^{k})$ is finite if and only if all moments of total order $k$ are finite.
