There is no difference between the formulas. The formula in the cyclotron radiation article is just the Larmor formula applied to the specific case of nonrelativistic charged particles moving in circles in a magnetic field.
For such particles, $F=ma$ combined with the Lorentz force law $F=qvB$ gives the acceleration as
$$a=\frac{qBv}{m}$$
so the Larmor formula
$$P=\frac{q^2a^2}{6\pi\epsilon_0c^3}$$
gives the radiated power as
$$P=\frac{q^2}{6\pi\epsilon_0c^3}\left(\frac{qBv}{m}\right)^2=\frac{q^4B^2v^2}{6\pi\epsilon_0m^2c^3}.$$
To see that this is the same as the cyclotron radiation formula,
$$-\frac{dE}{dt}=\frac{\sigma_tB^2v^2}{c\mu_0},$$
just substitute the formula for the Thomson cross section,
$$\sigma_t=\frac{8\pi}{3}\left(\frac{q^2}{4\pi\epsilon_0mc^2}\right)^2$$
and the formula for the vacuum permeability,
$$\mu_0=\frac{1}{\epsilon_0c^2}$$
to get
$$-\frac{dE}{dt}=\frac{8\pi}{3}\left(\frac{q^2}{4\pi\epsilon_0mc^2}\right)^2\frac{B^2v^2}{c}\epsilon_0c^2=\frac{q^4B^2v^2}{6\pi\epsilon_0m^2c^3}.$$
The total power $P$ given by the Larmor formula is the power radiated in all directions. The formula for the power per solid angle radiated in a specific direction is
$$\frac{dP}{d\Omega}=\frac{q^2a^2\sin^2\theta}{16\pi^2\epsilon_0c^3}.$$
Here $\theta$ is the angle between the acceleration vector and the direction in which the power is measured.
If you integrate this over all solid angles using $d\Omega=\sin\theta\,d\theta\,d\phi$, you'll get the Larmor formula for $P$ because
$$\sin^2\theta\,d\Omega=2\pi\int_0^\pi \sin^3\theta\,d\theta=2\pi\int_{-1}^1 (1-u^2)\,du=\frac{8\pi}{3}.$$$$\int\sin^2\theta\,d\Omega=2\pi\int_0^\pi \sin^3\theta\,d\theta=2\pi\int_{-1}^1 (1-u^2)\,du=\frac{8\pi}{3}.$$