Units of rate constant (cycles per second vs radians per second) in Eyring equation
I have looked at the original article of Henry Eyring (J. Chem. Phys. 1935, 3, 107). In that article, Eyring explains the derivation of his famous equation. It basically boils down to the following:

You assume an equilibrium between the initial and transition state and consider that once a species crosses the barrier, it goes to the product state:

$$R \leftrightarrows R^{\dagger} \rightarrow P$$

The rate expression is then simply:

$$k = \nu K$$

where $k$ is the reaction rate constant, $\nu$ the crossing frequency and $K$ the equilibrium constant between the IS and TS.

I will not go into detail about how to calculate $K$, but just remember that $K$ can be obtained by calculating the quotient of the partition functions of the system in the transition and initial state, respectively. How $\nu$ is calculated, will hopefully answer your question. 

Let us assume that in the transition state, the bonds to be formed or broken are weak. In that case, we can model the reaction coordinate (the direction of the reaction) as a loose vibration. The partition function of a vibration is

$$
f = \frac{1}{1 - \exp \left(\frac{-h \nu}{k_{b}T} \right)}
$$

Herein, $\nu$ is the frequency of the loose vibration. We assumed that this vibration was very weak (loose). In that case we can apply a Taylor expansion to the exponent and cut off the series after the first term. This gives us:

$$
f = \frac{1}{1 - 1 + \frac{h \nu}{k_{b}T}} = \frac{k_{b}T}{h \nu}
$$

Plugging this result back into our original rate expression gives

$$
k = \nu \frac{k_{b}T}{h \nu} K^{\dagger} = \frac{k_{b}T}{h} K^{\dagger}
$$

Here, I have put a $\dagger$ after the equilibrium constant to make clear that I have extracted one partition function (that of the loose vibration) out of the equilibrium constant.

From the dimensional analysis of $\frac{k_{b}T}{h}$, you can see that $k$ should have units of $s^{-1}$ and not something else.