Which reaction between Lugol and Hydrogen peroxide?
I'll try to figure out the color of every compound separately, but there is a strong interaction between them and the different parameters so this could be quite tricky:


$\ce{I+ , IO3-}$ don't absorb in the visible region
$\ce{I3-}$  I think should be red
$\ce{I2}$ is blue if solid but has different colors in solution depending from the type of solvent and $\ce{H+}$ concentration see Hildebrand (1909). In polar solvent is brown, however I don't exclude this is due to the interaction with $\ce{I3-}$.


Because there is an equilibrium:

$$ \ce{I_2 + I^- &lt;=&gt; I_3^-}$$

So there is always a little bit of iodine in the solution and this could cause the brown color (however keep in mind that is $I$ is not so soluble in water).
Adding $H_2O_2$ should oxidize the iodide in this way:
$$ \ce{2I^- + 2H^+ + H_2O_2 -&gt; I_2 + 2H_2O}$$
It occurs the reverse reaction too but should be negligible. In this case is probably a cinetic matter,  you can make the solution transparent reducing the colored compounds. If you have a look to the iodine clock reactions iodine can be reduced by bisulfite: 
$$ \ce{I_2 + HSO_3^- + H2O -&gt; 2I^- + HSO4^- +2H^+}$$

But I exclude that a reduction could occur in an oxidizing ambient like your. So my hypothesis are two:

FIRST: You are oxidizing the colored species. Adding $\ce{H2SO4}$ and $\ce{H2O2}$ you could obtain one of the most oxidant agent, Peroxymonosulfuric acid ($\ce{H2SO5}$),  see Piranha solution (I think this could be very dangerous please take all the precautions!). But now I don't think that the equation that if I wrote before (see J. Am. Chem. Soc., 1931, 53 (1), pp 38–44 and
J. Am. Chem. Soc., 1978, 100 (1), pp 87–91):

$$ \ce{I_2 + 5H2O2 -&gt; 2H^+ + 2IO3^- + 4H2O}$$

could occur. I know iodic acid $\ce{HIO3}$can be produced oxidating $\ce{I2}$ but I never found a reaction involving $\ce{H2SO4}$ and hydrogen peroxide. It could be a more complex reaction maybe involving the others ions. I was thinking about potassium hydrogen iodate ($\ce{KH(IO3)2}$). I try to figure out a stechiometric reaction but is quite an hell!

SECOND: simply you are  modifying the interaction between the solvent and $\ce{I2}$ and $\ce{I3-}$ avoiding the  light absorption to take place.

I'm sorry this is now quite not an answer, but I hope it helps!

Iodine color reaction are wonderful (but very complex)!