Electron Transfer in Organic Reactions: Why Does Brown Iron Oxide Improve Burn Rate of Rocket Candy?
First of all, you're thinking of one of the two definitions of oxidation—breaking bonds in molecular oxygen and rejoining them in a different way in another molecule.

We need the electron transfer definition of oxidation however. 

First off, to see what is happening we need to look at how many electrons are being transferred. In glucose the oxidation state of carbon is $\ce{0}$, whilst in carbon dioxide it is $\ce{+4}$. So during our rocket burning process, each carbon gives up four electrons. But what is it giving them to?

If we use molecular oxygen as our accepting species—our oxidisation agent—we have the following:

$$ \ce {6O_2} + \ce {C6H12O6}  \ce{-&gt;} \ce{6CO2} + \ce{6H2O} $$

With oxygen here going from an oxidation state of $\ce{0}$ in $\ce{O_2}$ to $\ce{-2}$ in both $\ce{H2O}$ and $\ce{CO2}$.

So if each $\ce{O2}$ molecule  is picking off two electrons and this is all it's doing, then why can't we replace this with two lots of $\ce{Fe^3+}$ to form $\ce{Fe^2+}$?

Before I write this in equation form, it's worth noting that organic electrochemical equations approximate organic matter (OM) with formaldehyde, $\ce{CH2O}$ since OM is biologically equivalent to a polymer (albeit complex ones!) of $\ce{CH2O}$. Note that glucose is therefore equivalent to six lots of this monomer. This simplifies writing balanced chemical equations a lot:

$$ \ce{2Fe_2O_3} + \ce {CH2O}  \ce{-&gt;} 4\ce{FeO} + \ce{H2O} + \ce{CO2} $$

Now molecular oxygen is by far a better oxidisation agent than $\ce{Fe_2O_3}$ however in fine powdered form, these iron oxide molecules are in immediate contact with your OM; we do not need to wait for molecular oxygen to be used up and for more to diffuse in.

So I'm warranting that red iron oxide ($\ce{Fe_2O_3}$) can be used as well as brown iron oxide (I think this means $\ce{FeO}$). The difference here is that in the former reaction each $\ce{Fe^3+}$ only captures one electron from your OM to form $\ce{Fe^2+}$, whilst in the latter case you're moving from $\ce{Fe^2+}$ to $\ce{Fe}$ metal which captures two electrons—so you need less. Note that formation of iron from $\ce{Fe^2+}$ will need really high temperatures or pressures, which is what you note in your question.

This is a super question in terms of getting people thinking about redox not just as a process which happens with ions, but with entire organic species. Such organic reactions can be characterised by a reaction voltage, just like the other types of electrochemical reaction you've been looking at. This is actually what happens in the field of biogeochemistry, where scientists look at which types of oxidisation agent are responsible for degrading OM in different environments. More can be found about this in this paper here.