One has $(F(1-G))'=(1-G)f-Fg$. So, for $F(1-G)$ to be increasing (that is, nondecreasing) in a right neighborhood (r.n.) of $0$, it is necessary and sufficient that $(1-G)f-Fg\ge0$ in a r.n. of $0$.
For the latter condition to hold, it is enough that
\begin{equation}
\liminf_{x\downarrow0}\frac{f(x)}{F(x)g(x)}>1, \tag{1}
\end{equation}
since $(1-G)f-Fg=[(1-G)\frac f{Fg}-1]Fg$ and $G(0+)=0$.
In turn, for (1) to hold, it is enough that one of the following conditions take place:
(i) $\liminf_{x\downarrow0}\frac{f(x)}{g(x)}>0$;
(ii) $f$ is increasing in a r.n. of $0$ and $\limsup_{x\to\infty}xg(x)<1$ for all $x$ in a r.n. of $0$;
(iii) $f$ is increasing in a r.n. of $0$ and $xg(x)$ is monotonic in $x$ in a r.n. of $0$;
(iv) $f$ is increasing in a r.n. of $0$ and $g$ is bounded in a r.n. of $0$.
Indeed, (i) implies (1), since $F(0+)=0$.
If (ii) holds, then for all $x$ in a r.n. of $0$ one has $F(x)\le\int_0^x f(t)\,dt\le f(x)x$ and hence $\frac{f(x)}{F(x)g(x)}\ge\frac1{xg(x)}$, so that (1) again holds.
Suppose now condition (iii) holds. Then $\lim_{x\downarrow0}xg(x)=:c\in[0,\infty]$ exists. If $c>0$, then for any $d\in(0,c)$ and all $x$ in a r.n. of $0$ one has $g(x)\ge\frac dx$ and hence $G(x)=\int_0^x g(t)\,dt=\infty$, a contradiction. So, $c=0$. Thus, (iii) implies (ii).
Also, clearly (iv) implies (ii).
Moreover, (iii) and (iv) can be considered pure regularity conditions.
Added: Let us also show the following: Under the apparently quite general ``non-oscillation'' regularity condition that $c:=\lim_{x\downarrow0}\frac{f(x)}{F(x)g(x)}$ exists in $[0,\infty]$ (which will follow if e.g. $\frac f{Fg}$ is monotonic in a r.n. of $0$), we must have $c=\infty$ and hence (1) will hold, and hence
$F(1-G)$ will be increasing in a r.n. of $0$.
Indeed, suppose that $c<\infty$. Then
\begin{equation}
\ln\frac{F(z)}{F(x)}=\int_x^z\frac{f(t)}{F(t)}\,dt=(c+o(1))\int_x^z g(t)\,dt
=(c+o(1))[G(z)-G(x)]\to0
\end{equation}
as $0<x<z\to0$ and hence $\frac{F(z)}{F(x)}\to1$ as $0<x<z\to0$. However, since $F(0+)=0$, for each $z>0$ we can always find some $x\in(0,z)$ such that $F(x)\le zF(z)$, which shows that $\limsup_{0<x<z\to0}\frac{F(z)}{F(x)}=\infty$. This contradiction completes the proof.
The monotonicity of $F(1-G)$ in a left neighborhood (l.n.) of $1$ can be considered quite similarly. Alternatively, it can be reduced to the monotonicity of $F(1-G)$ in a r.n. of $0$ by introducing distribution functions $F_1$ and $G_1$ defined by the conditions $F_1(y)=1-G(1-y)$ and $G_1(y)=1-F(1-y)$, so that for $x=1-y$ one has $F(x)(1-G(x))=F_1(y)(1-G_1(y))$ and $x\uparrow1\iff y\downarrow0$.