Yes. Without loss of generality, $X\geq 2$ is an integer. Then we have explicitly
$$ v(\beta)=e\left(\frac{(X+1)\beta}{2}\right)\frac{\sin(\pi X\beta)}{\sin(\pi\beta)},\qquad\beta\not\in\mathbb{Z}.$$
It follows for any $p>0$ that
$$\int_{|\beta|\leq\frac{1}{4X}}|v(\beta)|^p\,d\beta\asymp_p\int_{|\beta|\leq \frac{1}{4X}}X^p\,d\beta\asymp_p X^{p-1}.$$
Now we estimate the rest of the integral carefully.
On the one hand,
$$\int_{\frac{1}{4X}<|\beta|\leq\frac{1}{2}}|v(\beta)|^p\,d\beta\ll_p\int_{\frac{1}{4X}<|\beta|\leq \frac{1}{2}}|\beta|^{-p}\,d\beta\asymp_p \begin{cases}X^{p-1},&p>1;\\\log X,&p=1;\\1,&p<1.\end{cases}$$
On the other hand,
\begin{align*}\int_{\frac{1}{4X}<|\beta|\leq\frac{1}{2}}|v(\beta)|^p\,d\beta&\gg_p\sum_{1\leq m\leq\frac{X}{2}}\int_{\frac{1}{4}<m-|X\beta|<\frac{3}{4}}\frac{d\beta}{|\beta|^{p}}\\
&\asymp_p\sum_{1\leq m\leq\frac{X}{2}}\frac{1}{X}\left(\frac{X}{m}\right)^p\asymp_p\begin{cases}X^{p-1},&p>1;\\\log X,&p=1;\\1,&p<1.\end{cases}\end{align*}
Putting together these three bounds,
$$\int_{|\beta|\leq 1/2}|v(\beta)|^p\,d\beta\asymp_p
\begin{cases}X^{p-1},&p>1;\\\log X,&p=1;\\1,&p<1.\end{cases}$$
This also means that in your second display the correct exponent of $X$ is $2k-1$, not $k$.