$\newcommand{\al}{\alpha}\newcommand{\Ga}{\Gamma}\newcommand{\be}{\beta}\newcommand\ip[1]{\langle #1\rangle}\newcommand\R{\mathbb R}$This property is briefly proved in Section 5.1 of [this paper][1]. The proof is based on Basse's characterization of the spectral representation of Gaussian semimartingales, namely, [Theorem 4.6][2]. 

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Here are details on this. It follows from Basse's theorem that, if $(X_t)_{t\ge0}$ were a semimartingale, then we would have 
\begin{equation*}
	(t-s)^\al=g(s)+\int_s^t\Psi_r(s)\mu(dr) \tag{1}\label{1}
\end{equation*}
for all real $t\ge0$ and almost all (a.a.) $s\in[0,t]$, where $\al:=H-1/2\in(-1/2,0)$, $g\colon\R_+\to\R$ is square integrable on $[0,t]$ for all real $t\ge0$, $\mu$ is a Radon measure
on $\R_+$, and $\R_+\times\R_+\ni(t,s)\mapsto\Psi_t(s)\in\R$ is a measurable mapping such that $\|\Psi_r\|_{L^2(\R_+)}=1$. 

In view of the Tonelli theorem, it follows from \eqref{1} that for a.a. triples $(s,c,t)$ such that $0<s<c<t<\infty$ we have 
\begin{equation*}
	\int_c^t\al(r-s)^{\al-1}\,dr=\int_c^t\Psi_r(s)\mu(dr).  \tag{2}\label{2}
\end{equation*}
Since $\al(r-s)^{\al-1}<0$ for $r\in(c,t)$, we see that the Lebesgue measure on $(0,\infty)$ is absolutely continuous w.r.t. $\mu$, with some density $h$, so that $h(r)=\frac{dr}{\mu(dr)}$ for real $r>0$. So, \eqref{2} implies 
\begin{equation*}
	\int_c^t\al(r-s)^{\al-1}\,h(r)\mu(dr)=\int_c^t\Psi_r(s)\mu(dr)  
\end{equation*}
for a.a. triples $(s,c,t)$ such that $0<s<c<t<\infty$. So, 
\begin{equation*}
	\Psi_r(s)=\al(r-s)^{\al-1}\,h(r)  \tag{3}\label{3}
\end{equation*}
for $\mu$-a.a. pairs $(s,r)$ such that $0<s<r<\infty$. Since the Lebesgue measure on $(0,\infty)$ is absolutely continuous w.r.t. $\mu$, we get \eqref{3} for a.a. pairs $(s,r)$ such that $0<s<r<\infty$. 

For any real $r>0$ with $h(r)\ne0$, by \eqref{3},  
\begin{equation*}
	1=\|\Psi_r\|_{L^2(\R_+)}^2\ge\al^2 h(r)^2\int_0^r(r-s)^{2\al-2}\,ds=\infty,
\end{equation*}
a contradiction.

Finally, if $h(r)=0$ for a.a. real $r>0$, then, by \eqref{3}, $\Psi_r(s)=0$ for a.a. pairs $(s,r)$ such that $0<s<r<\infty$, which contradicts \eqref{1}. $\quad\Box$ 

  [1]: http://arxiv.org/abs/2012.00975v1
  [2]: https://link.springer.com/article/10.1007/s10959-009-0246-2