Your assumption is correct except for battery resistance and converter efficiency which increases your primary current and varies with cutoff voltage. This results in 50 to 60% error.
If you look up the Watt-hour rating vs time of discharge for Alkaline cells and examine the cutoff chosen for 3 cells matches or exceeds the lowest input range of your step-up converter.
You now have your Energy source =Wh x N cells
Then compute your load Power in Watts. since Energy is Power * time , the result is your energy draw time until cutoff voltage is reached. Step$up Voltage energy transfer efficiency might vary somewhat near cutoff since the battery resistance rises sharply.
Wh/h=h then times efficiency yields your estimated discharge time .
Normally battery ratings are done for 20h discharge periods, so reducing to near 1h and less leads to a loss of capacity due to the greater batteries ESR*I^2 losses which are heat and not recoverable. Thus is why Wh capacity may reduce to higher load currents.
If we assume the average voltage for a constant current you cells may have a 20 h energy capacity of 1.25VxAh=Wh (from 1.6 to 0.9V). we don’t know why you said “total” capacity because cells in series have an Ah capacity of the weakest cell, and NOT the total of all three. In any case your 900mAh seems low as AA alkaline may have 2.4Ah @25’C for 20h load which is reduced to less than half in 1h loads. this is due to the much higher ESR of Alkaline compared to Lithium and NiMH.
So if you see the 20h Energy capacity vs 1h capacity, you may have 3x energy capacity from voltages adding but current limited by the weakest cell.
Assuming your value is for a std 20h discharge rate compare 900mAh*2.25V=2.0Wh to 2.4Wh per cell on this chart and then later x3 for 3 cells.
Notice the 20h interpolation for the black curve of Alkaline AA intersects with 2.2Wh per cell and at 1h it is only 1.1Wh or 1.1W for 1h per cell. or 50% of rated capacity, which is coincidental to orange arrow.
It is not a simple solution, but it is more accurate.