No!  The CMB radiation is such a perfect fit to a blackbody that it
cannot be made by stars.  There are two reasons for this.

First, stars themselves are at best only pretty good blackbodies, and
the usual absorption lines and band edges make them pretty bad
blackbodies.  In order for a star to radiate at all, the outer layers
of the star must have a temperature gradient, with the outermost
layers of the star being the coolest and the temperature increasing
with depth inside the star.  Because of this temperature gradient, the
light we see is a mixture of radiation from the hotter lower levels
(blue) and the cooler outer levels (red).  When blackbodies with these
temperatures are mixed, the result is close to, but not exactly equal
to a blackbody.  The absorption lines in a star's spectrum further
distort its spectrum from a blackbody.
                                                                        
One might imagine that by having stars visible from different
redshifts that the absorption lines could become smoothed out.
However, these stars will be, in general, different temperature
blackbodies, and we've already seen from above that it is the mixing
of different apparent temperatures that causes the deviation from a
blackbody.  Hence more mixing will make things worse.
                                                                               
How does the Big Bang produce a nearly perfect blackbody CMB?  In the
Big Bang model there are no temperature gradients because the Universe
is homogeneous.  While the temperature varies with time, this
variation is exactly canceled by the redshift.  The apparent
temperature of radiation from redshift z is given by T(z)/(1+z), which
is equal to the CMB temperature T(CMB) for all redshifts that
contribute to the CMB.