EDIT:
Based on the mistake pointed below, I have computed the (now hopefully) correct metric and the differential equations for the geodesics.
I have the following co-ordinate transformation,
\begin{align} t & = t' \\ x' & = r \cos(\theta - \omega t) \\ y' & = r \cos(\theta - \omega t) \\ z' & = z \end{align}
Physically, this respresents rotations of co-ordinates, which can be most easily descrived in the cylindirical coordinate systems. If we drop the time dependent terms in the argument for the trignometric functions, the metric is that of the cylindirical polar coordinates,
\begin{align} ds^2 = -dt^2 + dr^2 + r^2 d \theta^2 + dz^2, \end{align}
where we have used the Minkowski metric to convert to metric tensor in the cylindrical polar coordinates. Now adding the time dependent terms, we move to the $(t', r', \theta', z')$ coordinates such that
\begin{align} t' & = t \\ r' & = r \\ \theta' & = \theta - \omega t \\ z' & = z, \end{align}
and
\begin{align} ds'^2 = (-1 + r^2 \omega^2)dt'^2 + dr'^2 + r'^2 d \theta'^2 + dz'^2 + r^2 \omega ( dt' d \theta' + d \theta' dt') .\end{align}
I now wish to calculate the most general solution of the geodesic. Using the only non vanishing Christoffel symbols
\begin{align} \Gamma^{r'}_{t' \theta'} & = \Gamma^{r'}_{\theta' t'} = -r \omega \\ \Gamma^{r'}_{\theta' \theta'} & = - r \\ \Gamma^{\theta'}_{t' r'} = \Gamma^{\theta'}_{r' t'} & = \frac{\omega}{r} \\ \Gamma^{\theta'}_{r' \theta'} = \Gamma^{\theta'}_{\theta' r'} & = \frac{1}{r}, \\ \end{align}
I get the following differential equations (where I have dropped the primes, except for the one on $\theta$):
\begin{align} \frac{d^2 t}{d \beta} & = 0 \\ \frac{d^2 r}{d \beta} - 2 r \omega \Bigg ( \frac{d t}{d \beta} \Bigg ) \Bigg ( \frac{d \theta'}{d \beta} \Bigg ) - r \Bigg ( \frac{d \theta'}{d \beta} \Bigg )^2 & = 0 \\ \frac{d^2 \theta'}{d \beta} + \frac{2 \omega}{r} \Bigg ( \frac{d t}{d \beta} \Bigg ) \Bigg ( \frac{d r}{d \beta} \Bigg ) + \frac{1}{r} \Bigg ( \frac{d r}{d \beta} \Bigg ) \Bigg ( \frac{d \theta'}{d \beta} \Bigg ) & = 0 \\ \frac{d^2 z}{d \beta^2} & = 0 \end{align}
Since $t' = t$ is an independent coordinate, we can choose $\beta = t$. From the last DE, we have $z = At + B$, and the second and third differential equations become
\begin{align} \frac{d^2 r}{d \beta} - 2 r \omega \Bigg ( \frac{d \theta'}{d \beta} \Bigg ) - r \Bigg ( \frac{d \theta'}{d \beta} \Bigg )^2 & = 0 \\ \frac{d^2 \theta'}{d \beta} + \frac{2 \omega}{r} \Bigg ( \frac{d r}{d \beta} \Bigg ) + \frac{1}{r} \Bigg ( \frac{d r}{d \beta} \Bigg ) \Bigg ( \frac{d \theta'}{d \beta} \Bigg ) & = 0 \end{align}
I am unsure how to solve these horrible looking coupled differential equations. I am not very good at DE's as of yet; it'd be great if someone could comment on my approach to the question (up till the point where I have calculated the DE's) and comment/hint on the difficulty of solving the equations. Or perhaps even answer the question after some comments have been exchanged.
In my experience, it is always worth looking for a Killing vector field on your spacetime rather than plunging headlong into the geodesic deviation equation. In your case, you have three such vector fields: one associated with time translations, one associated with rotations, and one associated with translations in the $z$-direction. This means that there are automatically three conserved quantities of the motion. Physically, these quantities will be related to the particle's energy, angular momentum, and $z$-momentum.
Once you have all three constants of the motion, I suspect you will be left with a single remaining ODE in terms of $r(\beta)$ and some of these constants. (You will probably also need to apply the condition that $v^a v_a = -1$. You can then find a formal solution (though you may not be able to express it in closed form) for $r(\beta)$.