To my knowledge, the $\ce{C-Cl}$ bond is stronger than the $\ce{C-Br}$ and the $\ce{C-I}$ bond.
I always thought that the stronger a bond, the harder it is to break so the boiling point would be higher.
If that is the case, why does, among $\ce{RCl}$, $\ce{RBr}$ and $\ce{RI}$ (R = alkyl), $\ce{RI}$ have the lowest boiling point?
Your knowledge of the bond strengths is correct, but your understanding of boiling is flawed.
As orthocresol has mentioned in his comment, boiling is not about breaking the carbon-halogen bonds in your tetrahalomethanes.
Boiling does not mean decomposition!
You only transfer your material from the liquid to the gaseous state. Here, the molecules are still intact. But upon heating to the the boiling point, you have overcome the interactions between the molecules that existed before and held the whole bunch of molecules together as a lquid.
So, why is there the observed trend in the boiling points? Did you have a look at the molecular weights? ;)
Let's have a look at the data for haloethanes, $\ce{C2H5-Hal}$:
$$\begin{array}{cr} \mathrm{Hal} &\mathrm{bp~[°C]}\\ \hline \ce{F} & -37 \\ \ce{Cl} & 13 \\ \ce{Br} & 38 \\ \ce{I} & 71 \\ \end{array}$$
Again, the trend is similar: Fluoroethane has the lowest, iodoethane has the highest boiling point.
Apparently, compounds with a higher molecular mass have a boiling point.
As a rule of thumb, the heavier the molecule, the more kinetical energy is needed to transfer it to the vapour phase. However, please note that this is a simplified model which might break when strong interactions between molecules, such as hydrogen bonds are involved!
