Must a sample of a deuterated compound be isotopically pure for kinetic isotope experiments?
There are two approaches to measuring kinetic isotope effects.

One is to have two completely separate, isotopically pure systems, one heavy, and one light. In each system you can measure the rate of the reaction however you want.  You don't have to be able to measure the rate in an "isotopically resolved" way, just measure the bulk rate.  Each of your systems is isotopically pure, so the rate of each system is 100% reflective of the isotope composition of that system.

The other way is to have a single system with variable/unknown isotopic composition.  You measure the rate of your reaction, but you have to be able to measure (e.g.) the rate of depletion of each substrate isotopomer separately.  (Or the rate of formation of each product isotopomer, but that sounds very difficult in your system.)  For example, if you had a mixture of $\ce{CH4}$ and $\ce{CD4}$ and you were interested in the kinetic isotope effect of some kind of combustion reaction of those gases with oxygen, you could measure the ratio of $\ce{CD4}$ to $\ce{CH4}$ over time, e.g. using mass spectrometry.  It would increase as the $\ce{CH4}$ is preferentially depleted by the combustion reaction.  You could also measure the rates of formation of $\ce{H2O}$ and $\ce{D2O}$.  The rate of formation of $\ce{H2O}$ would be much higher than that of $\ce{D2O}$.  (The exact numerical values here would depend strongly on the nature of your combustion system, e.g. temperature, catalysts used, etc.)

So iff you can measure $\ce{FeOOH}$ and $\ce{FeOOD}$ separately in the same system, then you don't need an isotopically pure material.  I'm not an expert in these types of measurements, but it sounds very hard to me.  Probably if you don't already know how to do it, it will be cheaper just to buy $\ce{KOD}$ and get on with the synthesis of an isotopically pure material so you can use the first approach.