I'm not sure why you think BJTs are significantly slower than power MOSFETs; that's certainly not an inherent characteristic. But there's nothing wrong with using FETs if that's what you prefer.
And MOSFET gates do indeed need significant amounts of current, especially if you want to switch them quickly, to charge and discharge the gate capacitance — sometimes up to a few amps! Your 10K gate resistors are going to significantly slow down your transitions. Normally, you'd use resistors of just 100Ω or so in series with the gates, for stability.
If you really want fast switching, you should use special-purpose gate-driver ICs between the PWM output of the MCU and the power MOSFETs. For example, International Rectifier has a wide range of driver chips, and there are versions that handle the details of the high-side drive for the P-channel FETs for you.
Additional:
How fast do you want the FETs to switch? Each time one switches on or off, it's going to dissipate a pulse of energy during the transition, and the shorter you can make this, the better. This pulse, multiplied by the PWM cycle frequency, is one component of the average power the FET needs to dissipate — often the dominant component. Other components include the on-state power (ID2 × RDS(ON) multiplied by the PWM duty cycle) and any energy dumped into the body diode in the off state.
One simple way to model the switching losses is to assume that the instantaneous power is roughly a triangular waveform whose peak is (VCC/2)×(ID/2) and whose base is equal to the transition time TRISE or TFALL. The area of these two triangles is the total switching energy dissipated during each full PWM cycle: (TRISE + TFALL) × VCC × ID / 8. Multiply this by the PWM cycle frequency to get the average switching-loss power.
The main thing that dominates the rise and fall times is how fast you can move the gate charge on and off the gate of the MOSFET. A typical medium-size MOSFET might have a total gate charge on the order of 50-100 nC. If you want to move that charge in, say, 1 µs, you need a gate driver capable of at least 50-100 mA. If you want it to switch twice as fast, you need twice the current.
If we plug in all the numbers for your design, we get: 12V × 3A
× 2µs / 8 × 32kHz = 0.288 W (per MOSFET). If we assume RDS(ON) of 20mΩ and a duty cycle of 50%, then the I2R losses will be 3A2 × 0.02Ω × 0.5 = 90 mW (again, per MOSFET). Together, the two active FETs at any given moment are going to be dissipating about 2/3 watt of power because of the switching.
Ultimately, it's a tradeoff between how efficient you want the circuit to be and how much effort you want to put into optimizing it.
Check out this device.
http://www.linear.com/product/LT1910
It's a high side N-channel MOSFET driver. I think that it has a little boost converter internally. We've just started playing with these and it will drive the MOSFET gate to 20 VDC with a 12 VDC supply. Nifty huh?!?
Though the device isn't fast, relatively speaking, so you would be better off doing PWM for speed control on the low side MOSFET. You can get a regular FET driver to do this with your 5 V input.
Best Answer
If this is a simple brushed DC motor with just two connections, then yes, you can tie one lead to power and use a low side switching element on the other. Don't forget to put a reverse Schottky diode accross the motor else the switch will get destroyed from the high voltage pulses.
At only 12 V, you can find N channel FETs that will switch well enough directly from a digital signal. These are often called "logicl level" FETs. In that case you don't need a FET driver.
Another possibility is a NPN as the switch. That is easy to drive from low voltage logic, but turning off a saturated bipolar is always a bit tricky. It depends on how fast you want to go. For a normal motor drive PWM frequency of about 25 kHz, it should be fine. However, unless you are using unusually low logic levels or the micro has a weak output drive, I'd look at logical level N channel FETs first.