That driver will work fine, provided you use a proper power supply.
The allegro stepper-drivers are current-limited chopper stepper drivers. As such, you only have to ensure the power-supply voltage for the driver is > then the rated voltage on the stepper, and you have set the current limit properly.
Basically, chopper-stepper-drivers actually modulate ("chop") the drive voltage to the stepper in real-time to maintain a fixed coil current.
The ratings for your motor are steady state. Basically, it says that if you apply 2.55V DC, 1.7A of current will flow though the motor coil.
However, the Allegro drivers don't apply DC, they apply a duty-cycle modulated square wave, which limits the overall power delivered to the motor.
Functionally, the driver will vary the applied voltage to the stepper to maintain a fixed current (it's not quite that simple, motor inductance is involved, but it's a reasonable simplification). As such, as long as you're not applying more then 1.7A of current to the motor, it will work fine.
Basically, the simple version is the motor ratings are basically constrained by the thermal behaviour of the motor. If you apply too much power, it'll get hot enough to damage the motor.
With the A4988 driver board you link, you can vary the motor current by adjusting the tiny pot, which allows you to adjust the motor power to whatever you'd like.
If you run the driver off input DC within it's operating range, you will be fine.
Based on what can be gleaned from the website you linked you ought not to exceed the phase current specified. This is how I see but if you can find proper data sheets, they might tell a different story.
Firstly look at the torque-speed curve (linked on the page): -
If you do the math, you'd calculate that the maximum mechanical output power is about 1 watt (100 rpm and 0.1 N.m = 1.047 watts). This is about the same at 200 rpm ( 200rpm and 0.06 N.m = 1.26 watts). At 400 rpm output power is 0.837 watts hence, you can see the max output power is 1.26 watts.
With a coil current of 0.31 amps per phase and a coil resistance of 38.5 ohms (as stated), the power (heat generated not mechanical power) is 0.31^2 x 38.5 watts = 3.7 watts and this means your stepper motor can be running quite hot.
Having said that, this "apparent" inefficiency (some simplification and assumptions made here) of about 25% will not be at optimum mechanical output. At optimum power output ( I reckon about 100rpm), the power in will be about 2.1 watts assuming a peak efficiency of about 60%. This is about "normal" for steppers of this type.
So, if you are always going to be running about 100 rpm, the current into the coils will be lower than 0.31 amps BUT, the details in the link are really unclear about this so caution should be taken.
Conclusion - I don't think you can dare run the coils at more that 0.31 amps based on what the specification says. I recommend finding out more about the device. Try looking at one from a regular supplier i.e. one that has a proper pdf data sheet and deciding what information that data sheet provides that this one doesn't.
Best Answer
Probably they will, if you check the "Absolute maximum ratings" in the product datasheet you see that the output currents are stated as "internally limited". Just be careful during the PCB design phase, 2.5A will generate quite a lot of heat. Be sure to design a proper copper plane to dissipate it, make sure to have large enough traces (planes) to carry current to the IC. Consider also adding an external heatsink on top of the chip, even if it has a plastic package it is better than nothing.
Anyway I am pretty sure 2.5A refers to the stall current (maximum current drawn, max torque, motor stalled), so you will probably stay under this current during normal operation of the motor.