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