The answer depends on the nature of the "huge oxygen bottles"...
You often see large ( around two stories high) tanks outside a hospital. They are distinguished by the name of the chemical firm, Union Carbide or Linde among others, painted on the side, and by the thick layer of frost found on the attached plumbing.
These tanks hold liquid oxygen at low pressure (a few atmospheres at most) and very low temperature with very effective(but cheap) insulation. The low amount of inward heat flow causes some liquid oxygen to vaporize, raising the internal pressure. The tank isn't very structurally strong, so the generated pressure is relieved by usage in the facility, or by a pressure relief valve, which of course wastes oxygen. In extreme cases, a heater may be used to raise the internal pressure to meet the needs of the facility. Failure of this pressure relief valve (closed) is a major danger for this type of tank.
The other type of oxygen tank is the much smaller (maybe 6 feet by 12 inch diameter) steel tank.
This is a very strong steel tank containing compressed oxygen gas at ambient temperature and very high pressure. The tank is heavy, expensive, and must be treated with great care. (I wouldn't use the set up in this photo; too tippy!) A complex pressure reducing valve is needed to give a controlled low pressure flow, If you examine such a tank (in a lab, for example), you may find a series of date stamps, possibly covering decades, on which the tank has been tested at some huge overpressure.
Note: image edited to eliminate problematic stamp on tank...
Surprisingly (it surprised me!) it doesn't make much difference. Well, if I do an approximate calculation it makes no difference but real life will differ a bit from my approximation. This is how it's done.
The cooling will be approximately described by Newton's law of cooling. This tells us that the rate of heat loss will be:
$$ W \approx kA\Delta T \tag{1} $$
where $\Delta T$ is the temperature difference, i.e. the temperature of the water minus the ambient temperature, $A$ is the surface area of the object and $k$ is some constant that has to be determined experimentally.
Take the case where you keep the boiling and cold water seperate. We'll use $T_h$ and $T_a$ for the temperature of the hot water and the ambient temperatures, and we'll assume the cold water is at ambient temperature. We'll take the mass of the hot water to be $M_h$ and the mass of the cold water to be $M_c$. The rate of heat loss is then simply:
$$ W_1 \approx kA(T_h - T_a) $$
Now suppose we mix the hot and cold water. We'll use $x_h$ to indicate the fraction of hot water i.e. the ratio:
$$ x_h = \frac{M_h}{M_h + M_c} $$
where $M_h$ and $M_c$ are the masses of the hot and cold water. The fraction of cold water $x_c$ is just equal to $1-x_h$. Now, the temperature of the mixture, $T_m$, will be:
$$ T_m = T_hx_h + T_ax_c = T_hx_h + T_a(1 - x_h) $$
The area $A$ changes as well. If you're pouring the water into a roughly cylindrical bottle the area will be proportional to the total mass of the water, so the area of the mixed water will be roughly:
$$ A_m \approx A \frac{M_h + M_c}{M_h} \approx A \frac{1}{x_h} $$
Now we have $T_m$ and $A_m$ we can plug them into equation (1) to get the rate of heat loss for the mixture:
$$\begin{align}
W_2 &= k A_m (T_m - T_a) \\
&= k A \frac{1}{x_h} \left( T_hx_h + T_a(1 - x_h) - T_a \right) \\
&= k A \frac{1}{x_h} \left( T_hx_h + T_a - T_ax_h - T_a \right) \\
&= k A \left( T_h - T_a \right) \\
&= W_1
\end{align}$$
and we end up with the surprising result that the two rates of heat loss are the same. This happens because although the hot water cools faster, the mixture has a greater surface area to cool through and the two effects cancel out.
Now, I did start out by saying this was just an approximation. Newton's law of cooling tends to fail at smaller temperature difference i.e. cold things cool more slowly that it would predict. So my guess is that keeping the water separate would actually cool it faster. However the difference probably won't be that great.
Best Answer
Yes, gas pressure affects the rate of heat transfer to or from a gas, but that's not what you're measuring.
What's probably going on:
If you've poured $200mL$ out of the bottle, the remaining liquid will cool faster, because the rate of heat transfer is roughly proportional to the surface area, while heat capacity is proportional to the volume. Surface area varies as the $2/3$ root of volume, so if you multiply the volume by a factor of $0.8$, you multiply the volume by $0.8^{2/3}$, and so the cooling time is multiplied by $0.8/0.8^{2/3} \approx 0.93$.
A pressure-related effect that is happening, but may be too small to measure:
When you open the soda, releasing the pressure, the dissolved carbon dioxide molecules boil out of the liquid. This is an endothermic process, so the liquid is cooled: heat flows from the liquid into the gas to pay for the phase change. I have no idea how much gas is released, but suppose it's somewhere on the order of 1 L from a 1L bottle being opened for a few seconds for the first time, which seems about right for how much they can froth over. Then that's about 2 g of CO2 , which has a latent heat of vaporization of about 700J, or about 1/5 of a degree worth of heat from your liter of water.
A pressure related effect that isn't happening:
As I said at the beginning - yes, gas pressure affects heat transfer to the gas. If you isothermally pressurized your freezer, the rate of cooling would increase because density increases the conductivity of air. This isn't what's happening here, of course, but it's a direct answer to your question so I didn't think I should leave it out. The effect on cooling rate for the kinds of pressures that wouldn't be life-threatening would be negligible.
If you actually want to cool the beverage quickly, just immerse it in a bucket of ice water. It'll cool rapidly to 0 and won't freeze.
EDIT - oops! I wrote exothermic when I meant endothermic. Fixed.