FFT padding to power of 2
It is possible but it is usually less complex to do the FFT of size n1.
To be honest given the FFT of a signal of size N1 padded to N2, the most efficient way that comes to my mind to find the coefficients of an FFT of the same signal without padding is to do the inverse FFT of the signal, removing the padding and do the FFT of size N1 with those clever libraries such as fftw.
Why?
Let $p(f) = \sum_{k=0}^{N_1-1} x_i \exp(2j \pi f k)$, the FFT of size $N_1$ is $p(n_1/N_1)$ but the FFT of size $N_2$ (with the padded signal) will give you $p(n_2/N_2)$.
So let's see for which $n_1$ we find solutions with integer $n_2$
$$\frac{n_1}{N_1} = \frac{n_2}{N_2}$$
it turns out that $n_2 = n_1 N_2 / N_1$, let $N_2 = 2^{a_2}$ and $N_1 = 2^{a_1} b$ for an odd $b$. Then we have $n_2 = n_1 2^{a_2 - a_1} / b$, i.e. only the coefficients at indices multiples of $b$.
In a more general way we could say that only the coefficients with indices multiples of $N_1 / \gcd(N_1, N_2)$ can be recovered.