= Solution
Let $\varphi=\phi-F(R)$, so
$$
d\phi=d\varphi+F'(R)dR.
$$
The new coefficient of $dR^2$ is
$$
-\frac{a^2\sin^2\theta}{\Delta}
+\Delta\sin^2\theta\,[F'(R)]^2.
$$
It vanishes if
$$
F'(R)=\frac a{R^2+a^2}.
$$
One choice that tends to zero as $R\to\infty$ and also has a regular $a\to0$ limit is
$$
\boxed{F(R)=-\arctan\frac aR}.
$$
Substitution gives the <outgoing null oblate spheroidal coordinates> form
$$
\boxed{
\begin{aligned}
ds^2={}&-du^2-2\,du\,dR
+2a\sin^2\theta\,dR\,d\varphi\\
&+(R^2+a^2\cos^2\theta)d\theta^2
+(R^2+a^2)\sin^2\theta\,d\varphi^2.
\end{aligned}}
$$
There is no $dR^2$ term, so $g_{RR}=0$ and $\partial_R$ is null for every $a\geq0$.
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