FFT of size not a power of 2
The resulting FFT will be different: instead of calculating the FFT at the frequencies $2\pi n/1800\ \ $ for $n=0\ldots 1799\ \ $, you will be calculating them at $2\pi n/2048\ \ $ for $n=0\ldots 2047\ \ $. However, there is no degradation of the information.

Both approaches are correct: using 1800 or 2048. I'd use the "minimum energy solution" (i.e. the simplest, the laziest solution).  This would usually be using the 2048 length transform.

People tend to use radix-2 transforms because they don't know any better. There seems to be much mis-information about FFTs HAVING to be power-of-2. There is no such constraint.  Also, they probably don't know about about decent non-radix-2 algorithms, such as those available in FFTW and other libraries.

To see that the FFT of any length is information-preserving:


  Assume your data of length 1800 is in x.  Form X = fft(x,2048); (or some other  length different from 1800). 
  Find xx = ifft(X);.
  See what sum(abs(x-xx(1:1800))); is.


See also this question and its answers.