In order to compensate for the $r$ in the volume element
$r\ dr\ d\phi\ dz$ you need a factor $\frac{1}{r}$ in 
your charge density $\rho$.
And the over-all constant is adjusted so that the total
volume integral will be $Q$. You finally get 
$$\rho(r,\phi,z)=\frac{Q}{2\pi rL}\delta(r)\theta(L/2-|z|)$$

The term $\frac{1}{2\pi r}\delta(r)$ might seem weird at first,
but it is actually equal to $\delta(x)\delta(y)$.
See also the math question *[Dirac delta in polar coordinates][1]*
and its accepted answer.


  [1]: https://math.stackexchange.com/questions/398777/dirac-delta-in-polar-coordinates