Two issues:
1. The USB-6009 has an internal pull-up resistor of 4700 ohms to +5 V. When power is first applied to the device (before any software has run), all the digital lines will be at 5 V. Thus, it is better to set the external circuit so that +5 V at V2 (the USB-6009 output) causes the motor to be off and 0 V turns the motor on. If the +/-12 V power supply can be on while the computer/USB hub is off, then you must be sure the motor will not run under that condition.
2. Unless the motor power is very low, it is probably better to switch the power at the +12 V and -12 V buses rather than ground. Shunt switching is not power efficient and relies on the switch transistors being kept on anytime the motor must be off. It is usually more difficult to design BJT circuits which are always on than always off.
I think this configuration may work. I have not checked it thoroughly for all possible modes. At V2 = +5 V Q1 conducts and all other transistors are off. At V2 = 0 V Q1 is off and the others conduct.
Lynn