voltage booster current draw
My input source is 3 x 1.5 V alkaline batteries.


OK. 4.5 V nominal supply.


  I used my multimeter to measure the current draw of my load and it reads 200 mAH.


Check your multimeter. It won't have a mAH range. You meant 200 mA.


  How long will my batteries last assuming they have a total capacity of 900 mAH?


The current drawn from the batteries will be scaled up by the step-up ratio divided by the efficiency. Let's assume 80% efficiency: \$ I_{batt} = \frac {12}{4.5}0.2 \frac {1}{0.8} = 0.67 \ \text A \$.

The batteries should last \$ \frac {990 \ \text {(mAh)}}{670 \ \text {(mA)}} =  1.5 \ \text {h}\$


  And will the current draw be the same if I used a 12 V power supply non boosted? (Assuming the voltage booster has an efficiency of 100%).


The current drawn at 12 V would be the same - but you won't find a 100% efficient booster.


  My thinking is that since the voltage is boosted, in order to maintain the same power rating, the current draw will increase, which makes the actual current draw 533 mAH in this case. 


Your calculation is correct (for 100% efficiency) but your units are wrong. 533 mA.