Analog controlled H-Bridge to digitally controlled

You don't want to control LEDs with voltage. It may appear to you that you get better control with voltage because a small voltage change causes a large apparent brightness. With current control, the LED brightness will be roughly linearly proportional to current. However, humans perceive brightness logarithmically which is why it may look to you like current control isn't working as expected.

The voltage to get a particular current and therefore brightness will vary between LEDs and also has a significant temperature dependency. You really want to control LEDs with current, not voltage.

You also seem to be asking about a linear control from a higher voltage such that the extra voltage times the LED current gets burned up as heat. PWM is used because it's more efficient. You design the circuit for reasonably constant current at the maximum the LED can handle, and have just enough voltage to guarantee this current. That means the system is pretty efficient at that max current. PWM then switches between that efficient on state and the off state, so the result is still efficient even when the current is half the maximum, for example.

However to answer your question, you can control the LED from a voltage that is low pass filtered from a PWM output. You have to put some kind of buffer or amplifier between the filtered PWM output and the LED. If I had to do it this way, I'd use a pass transistor driven by a opamp. The trick is to put a small low side current sense resistor between the LED and ground, and have the opamp make the voltage accross that resistor proportional to the filtered PWM signal. That will give you true current control.

What is the purpose of R2 in the first circuit? It's limiting the maximum current to the motor to 12V/220Ω = 54.5mA, which is probably not enough.

Just connect the motor directly to the collector of Q2.

You also need a couple of other changes: put a resistor (e.g., 4700Ω) between the MCU output and the base of Q1; otherwise, the B-E junction of Q1 looks pretty uch like a dead short to ground to the MCU. Similarly, you need a resistor (about 1200Ω) between the collector of Q1 and the base of Q2; without it, when Q1 switches on, it basically shorts the 12V supply to ground via the B-E junction of Q2. 1200Ω limits the base drive to Q2 to about 10mA, which should be plenty.