For simplicity, let's assume initially  that the support of $\mu$ is the whole real line, so that $F$ is a homeomorphism $ \mathbb{R} \to (0,1) $. Let's denote $b:=F(0)=\mu(-\infty,0]=1-\mu[0,+\infty)\in(0,1)$.  

The relevant function is the strictly convex function $\Phi :(0,1)\to\mathbb{R}$ defined as
$$\Phi(t):=\int_{b}^t F^{-1}(s)\, ds  $$
which is the Legendre transformation of $\int_0^x F(t)dt$. Incidentally  $\Phi(0)=-\int_{-\infty}^0 F(t)dt$,   $\Phi(1)=\int_0^{+\infty} (1 -F(t))dt$ and $\|\Phi\|_\infty=\max\{\Phi(0),\Phi(1)\}$ may be finite or not.

Let's denote $\Psi_{+}:=(\Phi_{|[b,1)})^{-1}:[0,\Phi(1))\to[b,1)$ and $\Psi_-:=(\Phi_{|(0,b]})^{-1}:[0,\Phi(0))\to(0,b]$ (they can be viewed as $C^1$ function with compact support on $[0,+\infty)$ by zero-extension). Note that $\Psi(t):=1-\Psi_+(t)+\Psi_-(t)=\big|\{\Psi\ge t\}\big|$ is the inverse function of the monotone decreasing rearrangement of $\Psi$, and is supported on $[0,\|\Phi\|_{\infty}]$.

Consider the solution $u:[0,+\infty)\to \mathbb{R}$ of the first order autonomous ODE 

$$u'(t) = \frac{1}{2} \Psi (u ),\quad t\ge0$$
with $u(0)=0$, given  explicitly by
$$u^{-1}(x)=2\int_0^x\frac{dy}{ \Psi (y) }$$


Then it is easy to check that $\phi_{\pm}:=F^{-1}\circ\Psi_\pm\circ u $  solve the initial problem written as
$$ 2\phi_+ F'(\phi_+)\phi_+'=2\phi_- F'(\phi_-)\phi_-'=1-F(\phi_+)+F(\phi_-)$$
that we can write equivalently, because $\phi_\pm=F^{-1}\circ F\circ \phi_\pm=\Phi'\circ F\circ \phi_\pm$, in the form
$$ 2(\Phi\circ F\circ \phi_+)'=2(\Phi\circ F\circ \phi_-)'=1-\Psi_+\circ\Phi\circ F\circ \phi_+  +\Psi_-\circ\Phi\circ F\circ \phi_- .$$ 

Conversely, if $\phi_\pm$ solve the initial system, the latter equation shows that $\Phi\circ F\circ \phi_+=\Phi\circ F\circ \phi_-$ because they coincide at $0$ and have the same derivative, and in fact solve the above equation for $u$.  

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**Rmk.** The assumption on the support of $\mu$ is not crucial; if it is not the whole real line, $F$ is constant on some open set, and $F^{-1}$ has jumps and it is only defined as a left inverse of $F$; the integral for $\Phi$ is however well defined.