We can actually do a pretty good estimation of which is better aerodynamically with some basic math.

Someone already noted the equation for drag:

Fd=1/2*pv^2*Cd*A

p and v are going to be constant for our little example here because we're driving the car in two different configurations through the same air at the same speed.

Now we need to fill in for some of the variables:

Cd of NC miatas is .38 for the NC1 and .36 for the NC2 according to "the internet". I'm not real up on my miata chassis, but again according to the internet a 2007 miata is an NC1, so we'll go with .38 for this example.

Some more internet sleuthing gives me a width of 67.7" and an overall height of 49". Let's subtract 5" of ground clearance to give the actual frontal area of:

AregularNCmiata = 67.7in*(49in-5in) = 2980in^2

Although the windshield certainly factors into the Cd, we're going to assume you keep roughly the same Cd when you chop the windshield, so we need to calculate the area without the windshield. Some quick MS Paint CAD showed that the windshield frontal area is ~1/3 of the overall height from the ground (49/3 = 16"). So the area without a windshield is

Anowindshield = 67.7in*(49in-5in-16in) = 1896in^2

You might already be seeing that the cage is going to need a LOT of area or a REALLY high drag coefficient to overcome that area reduction....

Luck for us, the drag coefficient of a circular shape is readily available (why, on the internet of course!). The Cd of a circular rod perpendicular to the wind is 1.2 in laminar flow (worst case scenario, it's .3 in turbulent flow).

We're going to assume no shadowing of the rear cage members, and the cage is roughly the dimensions of a windshield frame (two vertical supports the height of the windshield and one across). As a quick and dirty estimate, I'm going to say the cage has roughly 3 'windshield frames' worth of cage area. So I'm estimating the cage area as roughly:

Acage = ((2*16in)+67.7in)*1.75*3 = 523 in^2 (2 16 verticals plus the 67.7 width times a tube diameter of 1.75 time the roughly 3 'windshield frames' worth of cage)

So now to bring it all together:

Fd (regular miata) = .38 * 2980 = 1132 (not adding any units here because this is kind of a weird 'coefficient')

Fd (caged miata, no windshield) = .38 * 1896 + 1.2 * 523 = 1348

Huh, I wasn't expecting that. The caged one looks like it's about 20% worse than the windshielded one. That's actually pretty bad.

So let's talk for a minute about some of my assumptions. The areas I calculated for the car are probably fine. The Cd for the windshieldless car is probably less than the standard car, but I can't find any good numbers on that in a 20 second Google search. I'm pretty confident in my 'no-shadowing' assumption on the cage members, but as a general rule I'm guessing the car is going fast enough that the flow over the bars isn't laminar. In that case:

Fd (caged miata, no windshield, turbulent) = .38 * 1896 + .3 * 523 = 877

So if you're going fast enough to have turbulent flow over the cage (seems likely) the caged car is ~20% better.

After typing up the whole thing, I'm going to go with at track speeds the caged miata probably has less drag than the stock one.

But your gas mileage driving around the paddock is really going to suffer.