Removing the oxygen this way will remove much of the atmosphere
Earth's atmosphere is $5\times10^{18}$ kg. Nitrogen and Oxygen are the primary components and are approximately equal in mass. 20% of the atmosphere is oxygen by molecule, therefore about 20% of the mass of the atmosphere is oxygen, so there is $1\times10^{18}$ kg of Oxygen that needs to be removed.
There are many oxides of iron, and these oxides form from two different oxidation states of iron atom, iron (II) and iron (III). Let's assume a 2:3 ratio of Fe:O in the iron oxides that we would form, based on the multiple oxidation states of iron that we would dump in the atmosphere. The mass of three oxygen atoms is about 48 g per mol; while two irons is about 111 g per mol. So to react with $1\times10^{18}$ kg of Oxygen, we need at least $2\times10^{18}$.
Dropping this mass of iron onto a planet will convert the gravitational potential energy that this mass had relative to the planet into kinetic energy, released on the planet's surface as it impacts. Assuming the iron comes from outside the planet's orbit, the gravitational potential energy is equal to the energy required to escape the planet's gravity. In kinetic energy terms, we would plug the mass along with the planet's escape velocity into
$$KE = \frac{1}{2}mv^2.$$ The escape velocity of Earth is 11.2 km/s; the total energy imparted to the Earth by the falling iron would be over $1\times10^{26}$ J. Even if the iron came from an orbiting moon, the imparted energy would be at least 95% of this total (a moon that big can only be so close).
Going to the best wikipedia page in the world, we see that this is a problem. This is equivalent to about 20 years of solar energy striking the surface of the Earth, or 200 Chicxulub impacts. If you were going to do this over a millennia or two, that would be one thing, but if you want to do it in ten years, that is another.
How the atmosphere would be stripped
Adding enough iron to a planet to remove all the oxygen from the atmosphere in a 10 year period will be equivalent to hitting it with a dino-killing meteor every three weeks over that period. Since this kinetic energy is dissipated first in the uppermost parts of the atmosphere, the energy will be dissipated as heat in the upper parts of the atmosphere.
The total KE energy of the added iron would not be enough, by itself, to remove the entire atmosphere. But since the iron is added in dust form, it would be a safe assumption that all of its energy would be dissipated in the atmosphere, and most of that in the upper atmosphere.
The total energy addition, divided by 10 years and then by the surface area of the planet is
$$ \frac{1\times10^{26} \text{ J}}{3.2\times10^{8}\text{ s}\cdot5.1\times10^{14}\text{ m}^2} = 620 \frac{\text{W}}{\text{m}^2}.$$
At the top of the stratosphere, particle temperature is in the 270 K range, with a root mean square velocity around 500 m/s, meaning oxygen and nitrogen particles still need ~10.5 km/s of delta-v to escape.
But gaseous molecules are not all travelling the same speed. These particles' velocities are distributed according to the Maxwell distribution, which is itself a $\chi^2$ distribution with three degrees of freedom. From the chart in the last link, we can see that the $\chi^2$ value for $p=0.1$ with $k=3$ is 6.25 ($k$ is the degrees of freedom). This means that 10% of particles will have a 'value' of 6.25 in a Maxwell distributed set of particles. The mean of the distribution is $k$, which is three, and this is equivalent to the root mean square (rms) velocity of the particles. Thus, if the rms velocity of a group of particles is 3/6.25 = 0.48 times the escape velocity, then 10% of the particles will still be above escape velocity. This is a more likely explanation of how atmospheric escape would work.
In this case, the delta-v required to get 10% of the particles to escape is only 4.9 km/s, so the added iron is bringing 2 mols of atmosphere to this temperate, every second, over every square meter of Earth for 10 years.
Here is where simple math breaks down. The impact of heating in the last second will affect heating in the next second, there is some amount of heat loss through mixing, and other heat loss through radiation back into space. Furthermore, some particles will escape with very high velocities, carrying off a large amount of energy with them. But at a very simple level, the 2 mols per second per square meter is equivalent to $1\times10^{15}$ mols of atmosphere heated until 10% is at escape velocity every second. If there were no mixing or heat loss, this will bring the entire atmosphere to 10% escape in 45 hours.
How much will escape I really can't estimate with accuracy, so I will back off my previous claim the the entire atmosphere will be stripped. A significant portion will be stripped, but there is not enough kinetic energy in the falling iron for it all to be stripped.