= Solution
Because $u^0=c,dt/d\tau=\gamma c$ is constant, $dt/d\tau=\gamma$. The coordinate speed along the circle is consequently
$$
v=\frac{R|\omega|}{\gamma},
$$
so
$$
\boxed{R(v)=\frac{\gamma(v)v}{|\omega|}
=\frac{\gamma(v)mv}{|q|B}.}
$$
For the convention in the question, take $qB>0$, so $\omega>0$ and $\Delta\tau=2\pi/\omega$.
On the circular solution, the free part of the <relativistic charged-particle action> is
$$
S_{\rm free}=-mc^2\Delta\tau=-\frac{2\pi mc^2}{\omega}.
$$
In the gauge $A_y=Bx$, the interaction is the closed <line integral>
$$
S_{\rm int}=q\oint A_\mu dx^\mu
=qB\oint x\,dy.
$$
The orbit is clockwise for $qB>0$, so $\oint x\,dy=-\pi R^2$. Thus
$$
S_{\rm int}=-\pi qBR^2
=-\frac{\pi m\gamma^2v^2}{\omega}.
$$
Combining the two terms and using $\gamma^2v^2=c^2(\gamma^2-1)$ gives the <charged-particle action over one relativistic cyclotron orbit>
$$
\boxed{
S(v)=-\frac{\pi m}{\omega}\left(2c^2+\gamma(v)^2v^2\right)
=-\frac{\pi mc^2}{\omega}\left(1+\gamma(v)^2\right).
}
$$
Although the intermediate magnetic line integral depends on the chosen <magnetic vector potential>, the closed-orbit result is <gauge transformation>[gauge invariant].
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