This conjecture is true.
Indeed, for $h:=\ln(f_a-f_b)$ and real $x\ne0$ we have
\begin{equation}
h'(x)=R(x):=\frac{F(x)}{G(x)},
\end{equation}
where
\begin{equation}
F(x):=\sqrt{\pi }
\left(\text{erf}\left(\frac{x}{a}\right)-\text{erf}\left(\frac{x}{b}\right)\right),
\end{equation}
\begin{equation}
G(x):= a e^{-x^2/a^2}+\sqrt{\pi } x\,
\text{erf}\left(\frac{x}{a}\right)-b
e^{-x^2/b^2}-\sqrt{\pi } x \, \text{erf}\left(\frac{x}{b}\right);
\end{equation}
note that $G(0)=a-b>0$ and $G'(x)=\sqrt{\pi }
\left(\text{erf}\left(\frac{x}{a}\right)-\text{erf}\left(\frac{x}{b}\right)\right)>0$ for real $x>0$, so that $G>0$ on $[0,\infty)$.
Since the function $h$ is smooth and even, it is enough to show that $R$ is decreasing on $(0,\infty)$. In what follows, $a>b>0$ and $x>0$, unless otherwise indicated.
Let
\begin{equation}
R_1(x):=\frac{F'(x)}{G'(x)},
\end{equation}
\begin{equation}
R_2(x):=\frac{F''(x)}{G''(x)}
= -\frac{2 x \left(a^3 e^{x^2/a^2}-b^3
e^{x^2/b^2}\right)}{a^2 b^2 \left(a e^{x^2/a^2}-b e^{x^2/b^2}\right)};
\end{equation}
note that $R_2(x)$ is undefined at $x=x_{a,b}$, where
\begin{equation}
x_{a,b}:=a b \sqrt{\frac{\ln a-\ln b}{a^2-b^2}}.
\end{equation}
Then
\begin{equation}
\begin{aligned}
H(t)&:=R'_2(x)\frac{1}{2} a^3 b^3 \left(a e^{x^2/a^2}-b e^{x^2/b^2}\right)^2 \\
& =a^4 \left(b^2-2 t\right)-a b^5 e^{\left(\frac{1}{b^2}-\frac{1}{a^2}\right)t}+a^2 \left(b^4+4 b^2
t\right)-a^5 b e^{\left(\frac{1}{a^2}-\frac{1}{b^2}\right)t}-2 b^4 t,
\end{aligned}
\end{equation}
where $t:=x^2>0$, and
\begin{equation}
H'(t)\frac{ab}{a^2-b^2}=H_1(u):=\frac{a^4}{u}-2 a^3 b+2 a b^3-b^4 u,
\end{equation}
where $u:=e^{\left(\frac{1}{a^2}-\frac{1}{b^2}\right)t}>1$. The only root $u$ of the equation $H_1(u)=0$ that may be positive is
\begin{equation}
u_{a,b}:=\frac{a b^3-a^3 b+\sqrt{a^6 b^2-a^4 b^4+a^2 b^6}}{b^4}.
\end{equation}
Replacing now $e^{\left(\frac{1}{a^2}-\frac{1}{b^2}\right)t}$ and $e^{\left(\frac{1}{b^2}-\frac{1}{a^2}\right)t}$ in the expression for $H(t)$ by $u_{a,b}$ and $1/u_{a,b}$, respectively, we see that the condition $H'(t)=0$ for some real $t>0$ will imply
\begin{equation}
\begin{aligned}
H(t)&=a^4 \left(-\frac{b^4}{-a^2+\sqrt{a^4-a^2 b^2+b^4}+b^2}+2 b^2-2
t\right) \\
&-a^2 b^2 \left(\sqrt{a^4-a^2 b^2+b^4}-4 t\right)-2
b^4 t,
\end{aligned}
\end{equation}
which is a rather simple algebraic expression, which is actually $<0$ (still assuming $a>b>0$ and $t>0$).
So, $H(t)<0$ at any critical point $t>0$ of $H$.
Also, $H(0)=-a (a - b)^2 b (a^2 + a b + b^2)<0$ and $H(\infty-):=\lim_{t\to\infty}H(t)=-\infty$. So, $H(t)<0$ for all real $t\ge0$.
So, $R_2$ is decreasing on $(0,x_{a,b})$ and on $(x_{a,b},\infty)$.
Also, $R_1>0$ on $(0,x_{a,b})$, $R_1(x_{a,b})=0$, $R_1<0$ on $(x_{a,b},\infty)$, $R_1(0+)=\infty$, and $R_1(\infty-)=-\infty$.
So, by Table 1.1, $R_1$ is decreasing on $(0,\infty)$. Also, $R'(0)=-\frac2{ab}<0$ and $R(\infty-)=-\infty$. So, again by Table 1.1, $R$ is decreasing on $(0,\infty)$. $\quad\Box$
