Only if it's comparable in size to 25 mV.
Let's plug it into the Taylor series and see.
I(V) = exp(Vf/25mV) = I(V_0) exp(dV/25mV)
= I(V_0) * (1 + u + u**2/2! + u**3/3! + ....)
where u = dV/25mV. Taking just the normalized perturbation X,
X = I(V)/I(V_0)-1, we get
X = u + u**2/2! + u**3/3! + ....
If u = A sin(wt), then up to second order,
X = A sin(wt) + A**2(1-cos(2wt))/(2*2!) + ....)
where I've used the usual identity for sin**2.
All the odd order terms are at harmonics of w, but the even order terms also contribute DC, i.e. rectification. You can't get a DC component without an even-order nonlinearity, due to the behaviour of powers of sin(x).
In particular, the linear term A sin(wt) is just reproduced as an RF signal and doesn't do anything to the other things. (First order terms obey superposition, in other words.)
Thus the rectified contribution is
X_dc = A**2/4 + O(A**4).
(It really is quadratic for small signals.)
I'm entirely in agreement about your interference measurements themselves, of course, but it needs more RF than you think.
Cheers
Phil Hobbs