Sensor range and speed based on resistor
Without knowing the details of your device or your setup, it's clear that you are misreading the data sheet. Leakage current is simply the current you can expect when there is no light falling on the device - and if it weren't for your schematic there would be no way to know that it IS a phototransistor.

What you need to worry about instead is the current flow in the device when it is fully illuminated. Note that this should be a minimum number, and should take into account the fact that, over time, the optical path may accumulate dust and dirt. Once you determine this, using half to one third that value as a trigger level seems reasonable.

Then you can calculate your resistor value as you have done.

As it stands, your circuit is likely to be sensitive to any stray light, potentially including things like room lighting.

TL:DR - You probably want a much smaller resistor.

EDIT - With the phototransistor identified as a PT334-6C, the data sheet is informative. It identifies the collector current as about 2 mA with 1 mw/cm^2 of light shining on it. I don't know if your light source will actually produce this power, but let's assume that it is the case.

Giving yourself a 50% margin, the maximum current will be 1 mA. You do not identify the value of Vcc, but let's assume 5 volts. Then a 5k resistor will be an excellent choice. At full illumination the Schmitt trigger input will be about 0 volts (actually about 0.4 according to the data sheet). This should provide plenty of margin. The dark current, which you specified as 1 uA, will provide a 5 mV drop when the beam is blocked, so the worst-case dark voltage will be 4.995 volts.

Of course, all of this assumes that you can provide enough optical power. This may or may not be the case, and you do not seem to realize the necessity of calculating (or, better yet, measuring) it.