I think Dori's answer can be simplified using adjunction on $X$. Using Dori's notation, one has $2g(F)-2=K_XF=-b<0$, so the only possibility is that $b=2$ and the general $F$ has genus 0.

This already answers the original question, but, as Dori did, one can describe the fiber precisely. Imposing $LF=0$ one gets $a_2=2$. Now blow down $L$ to get a fibered surface $X'$. The image $D_2'$ of $D_2$ is a $-1$-curve, the images $D'_1$, $D'_3$ of $D_1$ and $D_3$ are again $-2$-curves and $2D'_2+a_1D'_1+a_3D'_3$ is a fiber $F'$ of a fibration of genus 0. Imposing $D'_2F'=0$ one gets $a_1D'_1D'_2+a_2D'_3D'_2=2$. Assuming $D'_1D'_2>0$, there are only two possibilities: (1) $a_1=a_3=D'_1D'_2=D'_3D'_2=1$; (2) $a_1D'_1D'_2=2$, $D'_2D'_3=0$, and in this case $D'_3D'_1>0$ since the fiber $F'$ is connected. Imposing $F'D'_3=F'D'_1=0$ it is easy to see that case (2) leads to a contradiction. So we have exactly the fiber described at the end of Dori's answer.

Finally note that a fibration with rational fibers cannot have a multiple fiber $F=mA$, since one would have $A^2=0$, $m=2$ and $K_XA=1$ but $A^2+K_XA$ is even, again by the adjunction formula.

I think Dori's answer can be simplified using adjunction on $X$. Using Dori's notation, one has $2g(F)-2=K_XF=-b<0$, so the only possibility is that $b=2$ and the general $F$ has genus 0.

This already answers the original question, but, as Dori did, one can describe the fiber precisely. Imposing $LF=0$ one gets $a_2=2$. Now blow down $L$ to get a fibered surface $X'$. The image $D_2'$ of $D_2$ is a $-1$-curve, the images $D'_1$, $D'_3$ of $D_1$ and $D_3$ are again $-2$-curves and $2D'_2+a_1D'_1+a_3D'_3$ is a fiber $F'$ of a fibration of genus 0. Imposing $D'_2F'=0$ one gets $a_1D'_1D'_2+a_2D'_3D'_2=2$. Assuming $D'_1D'_2>0$, there are only two possibilities: (1) $a_1=a_3=D'_1D'_2=D'_3D'_2=1$; (2) $a_1D'_1D'_2=2$, $D'_2D'_3=0$, and in this case $D'_3D'_1>0$ since the fiber $F'$ is connected. Imposing $F'D'_3=F'D'_1=0$ it is easy to see that case (2) leads to a contradiction. So we have exactly the fiber described at the end of Dori's answer.

Finally note that a fibration with rational fibers cannot have a multiple fiber $F=mA$, since one would have $A^2=0$, $m=2$ and $K_XA=-1$ but $A^2+K_XA$ is even, again by the adjunction formula.