Why boron trihydride is weaker lewis acid than boron tribromide and boron tri-iodide?
Predicting the acidic strengths of $\ce{BX_3}$ using back-bonding is one plausible theory. This article http://pubs.acs.org/doi/abs/10.1021/ic990713m hints us to use bond-strength data to predict the same:


  The weaker Lewis acid strength of $\ce{BF3}$ compared to $\ce{BCl3}$ with respect to bases such as $\ce{NMe3}$ is explained in terms of the ligand close-packing (LCP) model. The halogen ligands remain close-packed during the formation of the complex leading to an increase in the $\pu{B−Hal}$ bond length. Because a $\ce{B−F}$ bond is stronger than a $\ce{B−Cl}$ bond, more energy is required to stretch a $\ce{B−F}$ bond than a $\ce{B−Cl}$ bond. Hence $\ce{BF3}$ is a weaker acid than $\ce{BCl3}$.


Pulling up the bond strength data, I get:

Bond |  Strength (KJ/mole) 
--------------------------
B-H  |  389 
B-F  |  613 
B-Cl |  456 
B-Br |  377
B-I  |   ?


Building up on these two pieces of information, it can be said that since $\ce{B-H}$ bond is stronger than $\ce{B-Br}$ and $\ce{B-I}$ (see below), $\ce{BH3}$ is a weaker acid than $\ce{BBr3}$ and $\ce{BI3}$.

I'm afraid I couldn't find the data for $\ce{B-I}$ bond, but since we can compare the bond-strengths of boron trifluorides on the basis of back-bonding theory, we can be sure that the strength of $\ce{B-I}$ bond will be $\lt  \pu{377 kJ mol^-1}$.