$\newcommand{\ep}{\varepsilon}$This is at best a partial answer, with a simple result in the positive direction, which also highlights certain difficulties which may lead to a counterexample.
Let $p_+$ and $p_-$ denote, respectively, densities of the positive and negative parts of the Lebesgue--Stieltjes measure $df$, so that $p_+p_-=0$ almost everywhere (a.e.). The condition that the total variation of $f$ on any open interval $(a,b)$ with $0\le a<b\le1$ is $b-a$ means that $p_++p_-=1$ a.e. and hence a.e.
\begin{equation*}
p_+=1_S,\quad p_-=1_{S^c}
\end{equation*}
for some Legesgue-measurable set $S\subseteq[0,1]$, where $S^c:=[0,1]\setminus S$.
So,
\begin{equation*}
I:=\int_{[0,1]}g(x)\,df(x)=\int_0^1 h(x)\,dx,
\end{equation*}
where $g$ is any Riemann-integrable function on $[0,1]$ and
\begin{equation*}
h:=g1_S-g1_{S^c};
\end{equation*}
then, of course, $g$ is Lebesgue integrable and hence so is $h$, so that the integral $I$ exists.
Suppose now for a moment that, possibly up to the a.e.-equivalence, the density $p_+=1_S$ is Riemann integrable. Then, without loss of generality, $h$ is Riemann integrable as well. So, letting
\begin{equation*}
\ep_k:=
\begin{cases}
1&\text{ if }x_k\in S,\\
-1&\text{ if }x_k\in S^c,
\end{cases}\tag{1}
\end{equation*}
we get $\ep_k g(x_k)=h(x_k)$ and hence
\begin{equation*}
\frac1n\sum_{k=0}^{n-1}\ep_k g(x_k)
=\frac1n\sum_{k=0}^{n-1}h(x_k) \to\int_0^1 h(x)\,dx=\int_{[0,1]}g(x)\,df(x),
\end{equation*}
as desired.
This reasoning falls apart if $I_S$ is not a.e.-equivalent to a Riemann-integrable function -- which happens, for instance, when $S$ is the Smith--Volterra--Cantor (fat) set. Letting now $(x_j)$ be a specific, convenient equidistributed sequence with values in $S^c$ and probing a large enough collection of Riemann-integrable functions $g$, one might find a counterexample to the conjecture in general. Indeed, (1) strongly suggests that the $\ep_k$'s should mimic the set $S$, taking of course the membership of the equidistributed $x_k$'s in $S$ into account. This mimicking task may be too hard to accomplish if $I_S$ is not a.e.-equivalent to a Riemann-integrable function. Therefore, I think in general there is a counterexample -- but it would be an instructive surprise otherwise!
