$\newcommand{\al}{\alpha}\newcommand{\Ga}{\Gamma}\newcommand{\be}{\beta}\newcommand\ip[1]{\langle #1\rangle}\newcommand\R{\mathbb R}$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}. $\quad\Box$
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, $|T|=0$. $\quad\Box$