Just based on the quadratic drag of air, yes, the fired bullet would take longer to hit the ground.
Just consider the vertical force caused by the air friction:
$F_y = - F_{\rm drag} \sin \theta = - C (v_x^2 + v_y^2) \frac{v_y}{\sqrt{v_x^2 + v_y^2}} = - C v_y \sqrt{v_x^2 + v_y^2}$
Where $\theta$ is the angle above the horizon for the bullet's velocity, and $C$ is some kind of drag coefficient. Note that when the bullet is moving down $\theta$ is negative, as is $v_y$, so the overall vertical force is positive and keeps the bullet off the ground for slightly longer.
In the dropped case, $v_x = 0$, so we get $F_y = -C v_y^2$.
In the fired case, we can neglect $v_y$ in the radical (assuming it's much smaller than $v_x$) and we get $F_y \approx -C v_y |v_x|$.
In other words, the upward force on the fired bullet is stronger, by a factor of $v_x / v_y$.
So freshman-level physics is wrong, at least according to sophomore-level physics.
Bonus Case:
If you're assuming a flat surface on earth, it's worth considering that many "flat" things (like the ocean) actually curve down and drop off below the horizon. In case you want to account for this curvature, it may be worth going to the bullet's reference frame with $\hat{y}$ always defined to point away from the center of the earth. Note that this puts you in a rotating reference frame, and then look at the centrifugal "force":
$F_y = m a = m R \omega^2 = m R \left(\frac{v_x}{R}\right)^2 = m \frac{v_x^2}{R} $
Where $R$ is the radius of the earth and $m$ is the mass of the bullet. So again, an upward force, this time proportional to $v_x$ squared. Of course this is the same as pointing out that the earth curves away from a straight line, but it's another fun application of not-quite-freshman physics.
Now you can add in much more complicated aerodynamics, but there the question sort of looses its undergrad physics charm there and becomes an aerospace engineering question!
Vacuum doesn't suck air. In a vacuum it is the other air which pushes it into the empty space. Air like any other gas will expand to fill the volume.
So you would expect the atmosphere to spread out to fill the rest of the universe - and without gravity holding it onto the Earth, it would do.
Edit: Yes some air is continually lost. The molecules in the atmosphere are moving at a range of speeds, some of the very fastest ones will be moving fast enough to have enough energy to overcome gravity and escape. This is especially true for the lightest elements, eg. Helium, which move fast and feel the effect of gravity least.
Best Answer
One of the other rules of firearm safety is "Know your target and what is beyond it."
A firearm (of any sort) can be safely fired if there is nothing of value between the muzzle and the point where the projectile loses the last of its kinetic energy. Some of these safe places are called "firing ranges".
"Firearms, The Law, and Forensic Ballistics" by Margaret-Ann Armour provides formulae for the maximum range of a shotgun pellet based on the pellet diameter ($PD$):
$$r_\mathrm{yards} = 2200 \tfrac{\mathrm{yd}}{\mathrm{in.}} \times PD_\mathrm{inches}$$
or
$$r_\mathrm{meters} = 100 \tfrac{\mathrm{m}}{\mathrm{mm}} \times PD_\mathrm{millimeters}$$
When shooting skeet as in your question, you might select a shot between .110 inch (2.79 mm) and .080 inch (2.03 mm). Doing the math, we come up with approximately 175-275 meters total projectile distance. That range (of values) is the maximum range (in meters) you need to make a (firing) range...and most shots will fall harmlessly to Earth much, much closer than the maximum.
You can see this when you look at a satellite image of a shotgun range. There is nothing of value within 300 meters of the firing line, but most of the shot falls to Earth within 50m.
Oh, and just to drive home the point that we are really, really certain there won't be any risk beyond those distances, what could that be in the top-left corner of the image?
It's an airport!