Why is fluorine more reactive than iodine despite the weaker I-I bond?
In comparing the reactivity of fluorine or iodine, there are a couple of factors to consider. Part of it is the strength of the $\ce{X–X}$ bond, which is a barrier that tends to discourage them from reacting. Obviously, the higher this barrier, the less reactive the halogen is (in general).

However, unless you're trying to consider the relative reactivity towards homolytic cleavage of the bond, it isn't sufficient to look at the bond strength. The other bit is the driving force for the formation of the products – in other words, something that encourages them to react.

In this case, the formation of fluorine-containing products is generally much more thermodynamically favourable than that of the corresponding iodine-containing products. Some factors influencing this are:


The electron affinity of fluorine is much larger, which makes reduction to fluoride much easier
$\ce{E–F}$ bonds are stronger than $\ce{E–I}$ bonds, in both an ionic and covalent sense. 
Fluoride has a larger solvation enthalpy than iodide


For example, factors 1 and 3 lead to the large differences in the reduction potentials: $E^\circ(\ce{F2}/\ce{F-}) = \pu{+2.87 V}$ versus $E^\circ(\ce{I2}/\ce{I-}) = \pu{+0.54 V}$.

Silica is readily attacked by $\ce{HF}$ not because it is a strong acid, but rather because of the strength of $\ce{Si-F}$ bond thus formed.

That deals with the thermodynamic parts of "reactivity", in the sense that it explains why some reactions with iodine require elevated temperatures for any significant conversion, whereas reactions with fluorine proceed readily at room temperature.

The same factors, though, apply to the kinetic aspects of "reactivity". The activation energy is largely dictated by bonds that one has to break or form, for example, and the rate of electron transfer processes usually increases with $E^\circ$. Exceptions to this arise when – for example – the reactants have solid state structures which must be broken apart before reaction can occur. However, that is not the case here.