Turning off external sensors with Arduino: MOSFET vs I/O pin
The way you powered your sensors is exactly correct, using an external FET to switch the supply.
I'd almost bet that the difference you see in the current is due to some capacitance in the sensors. 

In both situations I am assuming you actually use an I/O pin on the Arduino:


Using the external FET you are driving the FET gate from the Arduino pin. The gate does have some capacitance, but you'd be hard pressed to show the voltage rise on the pin without an Oscilloscope, and even more challenged to accurately measure the current from the I/O pin in charging the gate. This gate capacitance is why you see the series resistor added from the I/O pin to the gate ….to limit the maximum current drawn (you sacrifice turn on time). When the FET turns on, it is capable of delivering much more current from the VCC supply than the I/O pin, so charges the sensor capacitors very much more rapidly.  The high current (it may actually be 10-100mA) during which the sensors supply voltage increases will be very short.  Without an Oscilloscope you will never see this effect.
When powering sensors directly from the I/O pin you have to directly charge any capacitors in the sensors (could be anything from 0.1-100uf depending on the sensor) via the I/O pin. To charge this capacitance you have to draw a lot of current from the I/O pin. The I/O pin exceeds its current rating and the voltage drops, which simply increases the time taken to charge the sensors capacitors.
This allows you to see the current flow for much longer, and it's easier to see on a multimeter (I assume this is what you did).   


Powering sensors directly from the I/O pins might in the extreme pop the fuse for that I/O block in the MCU. I'd advise against it unless you really understand your sensor circuitry.    

For very low current functions such as a real time clock, you will see them controlled directly from an I/O pin. Again, providing you know exactly what the function/sensor circuit is you are taking a risk. 

Update:

You don't describe your sensors in any detail, but the capacitance I'm talking about is that which would be used on the supply line:







simulate this circuit &ndash; Schematic created using CircuitLab