$\newcommand{\al}{\alpha}\newcommand{\Ga}{\Gamma}\newcommand{\be}{\beta}\newcommand\ip[1]{\langle #1\rangle}\newcommand\R{\mathbb R}$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
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**Details on \eqref{2}:** Recall that \eqref{1} holds for all real $t\ge0$ and a.a. $s\in[0,t]$. Replace each of the two entries of $t$ in \eqref{1} by $c\in(s,t)$ and refer to this modification of \eqref{1} as \eqref{1}$_c$. By the Tonelli theorem, the conjunction of \eqref{1} and \eqref{1}$_c$ holds for a.a. triples $(s,c,t)$ such that $0<s<c<t<\infty$. For any such $(s,c,t)$, subtract the left-hand side of \eqref{1}$_c$ from that of \eqref{1}, and also subtract the right-hand side of \eqref{1}$_c$ from that of \eqref{1}. Then ($g(s)$ disappears and) we get \eqref{2} for a.a. triples $(s,c,t)$ such that $0<s<c<t<\infty$.
**Further details, on the use of the Tonelli theorem, to show that the conjunction of \eqref{1} and \eqref{1}$_c$ holds for a.a. triples $(s,c,t)$ such that $0<s<c<t<\infty$:** Let $T\subset\R^3$ denote the set of triples $(s,c,t)$ such that $0<s<c<t<\infty$ and the conjunction of \eqref{1} and \eqref{1}$_c$ fails to hold. We want to show that the Lebesgue measure $|T|$ of $T$ is $0$. Let $S\subset\R^2$ denote the set of pairs $(s,t)$ such that $t\ge0$, $s\in[0,t]$, and \eqref{1} fails to hold. Then, in view of the Tonelli theorem,
\begin{equation*}
\begin{aligned}
|S|&=\iint_{\R^2}ds\,dt\,1(t\ge0, s\in[0,t], \text{\eqref{1} fails to hold}) \\
&=\int_\R dt\,1(t\ge0)\int_\R ds\,1(s\in[0,t], \text{\eqref{1} fails to hold}) \\
&=\int_\R dt\,1(t\ge0)0=0,
\end{aligned}
\end{equation*}
since \eqref{1} holds for all real $t\ge0$ and a.a. $s\in[0,t]$.
Note that
\begin{equation*}
T\subseteq\{(s,c,t)\colon(s,t)\in S\text{ or }(s,c)\in S\}.
\end{equation*}
So, again in view of the Tonelli theorem,
\begin{equation*}
\begin{aligned}
|T|&=\iiint_{\R^3}ds\,dc\,dt\,(1(s,c,t)\in T)) \\
&\le\iiint_{\R^3}ds\,dc\,dt\,(1((s,t)\in S)+1((s,c)\in S)) \\
&=\int_\R dc\,\iint_{\R^2}ds\,dt\,(1((s,t)\in S)
+\int_\R dt\,\iint_{\R^2}ds\,dc\,(1((s,c)\in S) \\
&=\int_\R dc\,|S|
+\int_\R dt\,|S|=0.
\end{aligned}
\end{equation*}
Thus, $|T|=0$. $\quad\Box$