How does doubling the voltage amplitude change the decibels?
If the voltage is increased by G then the signal has increased by: -

\$20\cdot log_{10}(G)\$ decibels.

This would also mean the same decibel increase in SPL. However, for "A" weighting you should really take account of the actual SPL level before making comparisons. A quote from wiki says this: -


  A-weighting is only really valid for relatively quiet sounds and for
  pure tones as it is based on the 40-phon Fletcher–Munson curves which
  represented an early determination of the equal-loudness contour for
  human hearing.


The reason is due to non-linearities in the ear and is typicfied by the above-mentioned fletcher Munson curve: -



But history has overtaken the FM curves and the equal loudness contours (shown in red above) are a better fit for what the ear does.

So, if your question was more aligned to understanding loudness as perceived by the ear, you must take into account the actual sound level and apply the more appropriate equal loudness curve.

EDIT to service extra question and comment

If a signal voltage rises by 6 dB then the power also increases by 6 dB and the sound pressure level also increases by 6 dB. It so happens that to increase the voltage by 6 dB you must double it and if the power increases by 6 dB it is quadrupled. This is because power is proportional to voltage squared hence a doubling of voltage (or pressure) results in four times the power.

So we often say a doubling of power is 3 dB and if it is doubled twice (i.e. made 4x bigger) it increases by 6 dB. Note that 6 dB is a slight approximation to the real figue i.e.: -

\$20\cdot log_{10}(2)\$ = more closely 6.0205 dB and

\$10\cdot log_{10}(4)\$ = the same 6.0205 dB

This is because \$log_{10}(x^2)\$ = \$2\cdot log_{10}(x)\$