Why is this intermediate borane-hydroperoxide adduct unstable?
Basically an intermediate form, that fulfills the octet rule, would be more stable than the trialkylborane, but only if we suppose that there are no other effects! And if there were no other ffects, it wouldn't be an intermediate form but an isolable compound.
In your reaction scheme you see that $\ce{OH-}$ will leave. You maybe know that $\ce{OH-}$ is - under basic conditions - a good leaving group. Moreover a peroxide is mostly pretty unstable, so it's happy to react and let $\ce{OH-}$ leave. The hydroboration–oxidation reaction gives you the tetrahydroxyborane-anion ($\ce{B(OH)4-}$) and three alcohols ($\ce{ROH}$) as final product. So you'll get an anion that fulfills the octet rule and is more stable than your reactant.

Edit: More details on the hydroxide as leaving group
The reaction finds place under strongly basic conditions. Usually you take $\ce{NaOH}$ to deprotonate the $\ce{H2O2}$ and form $\ce{OOH-}$, so in your solution there are plenty of $\ce{OH-}$ ions. The intermediate also splits off an hydroxide ion which is a good leaving group under this conditions. This means that the ion is stabilized in solution and is not less stable as the intermediate. Even if you can't really compare the stability of those two things because they are not the same... So let's only say, that the hydroxide ion is pretty stable in a basic solution.
As I also said in my original answer above, peroxides are pretty unstable. This is because the two oxygens are right next to each other. As you may know, oxygen is one of the most electronegative elements of the periodic table, so it &quot;fights to get electrons&quot;. If you connect two oxygens like in a peroxide, they both fight to get the electron of the other one, so it resembles rope pulling. This isn't the best case scenario for a stable bond.
If you look at hydrogen peroxide, especially in concentrated solutions, it tends to decompose without any external influence.
$\ce{2H2O2 -&gt; 2H2O + O2}$
This reaction releases 98,2 kJ/mol [1] of energy. This means that the products (oxygen and water) are way more stable than the hydrogen peroxide itself. This is mainly due to the very unstable rope pulling-bond between the two oxygens in $\ce{H2O2}$: $\ce{RO-OR}$.
This should explain why the intermediate group is so unstable.