Well, yes if you actually take the amplitude over one cycle then it will be zero for a sinusoid. But for sinusoidal signals, the average value is calculated over half a cycle. That was one point I missed in my earlier reply, and the squaring part .
However, I believe there is a small correction to technique you outlined. We will get instantaneous power at each sample point and integrating them will not yield the correct value. For example, for every sample, the power will never be negative and if you simply accumulate them the resulting value will only increase. Let's say, at t = 0, v = 1V, at t = 2, v =1V, at t = 3, v = 3V, at t = 4, v = -1V. Same thing repeats for t = 5 to t = 9. Note that this is a non-sinusoidal but repetitive signal.
Squaring and accumulating for t=0 to 4, we have
0^2 + 1^2 + 2^2 + 3^2 + (-1)^2 = 0 + 1 + 4 + 9 + 1 = 15
taking the square root, we have, 3.87V
Squaring and accumulating for t = 0 to 9, we have
0^2 + 1^2 + 2^2 + 3^2 + (-1)^2 + 0^2 + 1^2 + 2^2 + 3^2 + (-1)^2 = 30
taking the square root, we have, 5.47V
The above does not look correct. However when we average the squares,
For t =0 to 4, we have 15/5 = 3 (sq. root = 1.732)
For t = 0 to 9 also, we have 30/10 = 3 (sq. root = 1.732, as above)
And this looks much more plausible.