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There won't be much of an effect, if at all.

Note that this is an order of magnitude calculation. I'll be making approximations/assumptions left and right.

The drag force will be $\frac12\rho v^2c_D A$, and by taking some approximations (as there are too many variables) we get $\frac12 1.2 v^2 0.1 (\pi 0.05)^2 =0.00047v^2$. This can produce a torque up to $F\frac{L}2=0.0056v^2$, which in turn produces an angular acceleration of $\alpha=\frac\tau{m}=6\times 10^{-8}v^2 \:\mathrm{s^{-2}}$

Let's say that the rocket accelerates constantly till it leaves the bulk of the atmosphere (after which point the drag doesn't apply). I'm taking the acceleration as 17 from the thrust values.

Then,$\alpha(t)=6\times 10^{-8}v^2*17*17t^2\implies \theta(t)=6\times 10^{-8}17*17 frac{t^4}{12}$. For the first 10 seconds (when this constant-acceleration scheme holds --after this its velocity stabilizes and the atmosphere is thinner, so the net deviation should be of the same order of magnitude or less), this gives us a deviation of 0.01 radians. The total deviation will be of a similar order of magnitude.

This isn't much. Winds can generate a much larger deviation. Of course, the deviation of a couple hundredths of a radian is a lot when talking about the moon, but again, the deviation due to other effects is a lot more, so the onboard stabilizers can take care of this pretty well.

Values used in answer: Mass of rocket: $90,000\:\mathrm{kg}$, Acceleration: $17.8 \:\mathrm{m/s^2}$, Length: 24 m, Drag coefficient of frog: To the order 0.1, Radius of frog: $0.05 \:\mathrm{m}$. h/t @TildalWave for providing this info :)