Can Doppler shift make colors of two stars at same distance different?
As you correctly mention, there are rare stars which move particularly fast. Check also hypervelocity stars section here, where a stellar population moving at around $10^3 \textrm{km/s}$ is described (by the way, these velocities were most likely produced in the galactic center). Note also, that these stars are too fast to stay in the disk for long times. As another sidenote, galactic center in not accessible in visual due to dust extinction.

Redshifts are calculated particularly easy: 
$$z=\dfrac{v}{c}$$, 
where $v$ is the velocity along the line of sight, $c$ is the speed of light, $z=\dfrac{\delta\lambda}{\lambda}$ and $(v/c)^2\approx 0$ (which holds even for hypervelocity stars).

Because hypervelocity stars are rare, they most likely can be found only close to regular stars. Yet, being very optimistic, take $v=3\cdot 10^3 \textrm{km/s}$, and you obtain that $\dfrac{\delta\lambda}{\lambda}=10^{-2}$ under very favourable conditions. In general it is not small, but when it comes to colors, the change is negligible.

Say, green light at $\lambda=555 \textrm{nm}$ would now have $\lambda=550 \textrm{nm}$ or $560 \textrm{nm}$ in wavelegth, i.e. would still be very green. So Doppler effect has no bearing on your problem. 

What has possible bearing, however, is dust reddening. One of the two stars might be obscured by the dust, unlike the other one, and look reddish. So, both b) and c) are possible answers.

P.S. And actually, if a star "appears reddish", it is most likely dust extinction. Black body temperature colors are either blue, bluish, white, yelowish, yellow, orange or red.