Rate of hydrolysis of silicon tetrahalides
We can justify the rate of hydrolysis of Silicon tetrahalides if we just think on the statement carefully. The statement claims that the rate of hydrolysis is in the order $\ce{SiF4 &gt; SiCl4 &gt; SiBr4 &gt; SiI4}$. So, we are mainly concerned about the rate of the reaction, not on the thermodynamic stability or any energetic criterion for this reaction( I have dealt with it in the next paragraph). If we recall the characteristics for rate of   a reaction, it depends on the slowest step of the reaction. In the hydrolysis reaction of silicon tetrahalides, the first step, i.e the attack of a $\ce{H2O}$ molecule to form the trigonal bipyramidal intermidiate is the rate determining step. Also note that, no cleavage of $\ce{Si-X}$ bond happens in the rate determining step, but it actually happens in the next step, which is a fast step, so it doesn't contribute to the rate of the reaction. So, basically strength of $\ce{Si-X }$ bond doesn't affect the rate of hydrolysis reaction (the situation is exactly same as Nuleophilic aromatic substitution reactions ($S_NAr_2$) also). But if the $\ce{Si}$ atom becomes more and more electrophilic, the attack by a water molecule becomes more easier which subsequently results in the increase of the rate of the reaction, So the electrophilicity of $\ce{Si}$ in the tetrahalides increase in the order $\ce{SiF4 &gt; SiCl4 &gt; SiBr4 &gt; SiI4}$, and so the rate of the hydrolysis also. 

Here is the mechanism of the first out of the four subsequent substitutions in the hydrolysis reaction of Silicon tetrahalides. 



Secondly, if we really want to judge the rate of the reaction by numerical calculations, observe that the ultimate product of hydrolysis reaction of all the tetrahalides are same. Also, all the reactions are exothermic in nature. So, $\Delta _r H^0$ (assuming the  hydrolysis reactions are done in standard conditions; it's not a strict assumption, it may not be at standard conditions also, but the changes in the heat of reaction will be almost similar for all the tetrahalides) will be more if $\Delta_f H^0$ is more for the reactants (note that, $\Delta _r H^0 = \Delta_f H^0_{products} -\Delta_f H^0_{reactants} $ and we are bothered about reactants. If $\Delta_f H^0_{reacants}$ are more $\Delta_r H^0$ will be more negative and hence energetically favourable). The values for $\Delta_f H^0$ have the order, $\ce{\Delta_fH^0_{SiF4} &lt; \Delta_fH^0_{SiCl4} &lt; \Delta_fH^0_{SiBr4} &lt; \Delta_fH^0_{SiI4}}$. So, the hydrolysis reaction of $\ce{SiF4}$ will be the least exothermic and that of $\ce{SiI4}$ will be the most exothermic, which is expected also from the bond energy considerations. So, if you have less release of heat energy, i.e thermodynamically less stable product, it will have lesser activation energy and higher rate constant, and thus the rate of the reaction will be highest for $\ce{SiF4}$ and least for $\ce{SiI4}$ as we will get the product along with relatively more thermodynamic stability. So, by thermodynamic consideration also, rate of hydrolysis should be, $\ce{SiF4 &gt; SiCl4 &gt; SiBr4 &gt; SiI4}$. 

Lastly, the $\pi$-bonding is not important in case of large size difference between $\ce{Si}$ and $\ce{F}$ ($2p$ and $3d$ have course greater size difference though you may not admit). The fluorine atom makes the silicon more electrophilic by its electron pulling ability rather than  back-bonding.