Can a rocket igniter rated for 12v and 3 amps be powered with a 4v battery at 9 amps?
Using Ohm's law you can modify the power formula.

$$ P = V I = (IR)I = I^2R \tag 1$$
$$ P = V I = V \frac {V}{R} = \frac {V^2}{R} \tag 2 $$

Since your R is constant then dropping the voltage will reduce the power.


  I have a rocket motor igniter that says it requires a 12 volt battery at 3 amps.


We can calculate the igniter resistance as \$ R = \frac{V}{I} = \frac {12}{3} = 4 \ \Omega \$. From (1) above we see that \$ P = \frac {V^2}{R} = \frac {12^2}{4} = 36 \ \text W \$. 

At 4 V the power dissipated would be \$ P = \frac {V^2}{R} = \frac {4^2}{4} = 4 \ \text W \$. Note the 'square' relationship: dividing the voltage by 3 divides the power by 9.