If your angle of attack in landing pattern is 14 degrees but your vertical velocity is 600 feet/minute, does the HUD show you 1G or not? Why?
I presume, an implicit assumption in your question is that we are flying a perfect glideslope, that is, a straight line with a constant speed, and without disturbances.
If this is the case, the acceleration will be zero by definition, and the only effect we (and an accelerometer) will feel is gravity.
The 'G-meter' shows the acceleration (incl. gravity) with respect to a calibrated standard 1G. Now, the question is, will gravity be the same from the point of view of the device/pilot?
The answer is: almost, but not quite.
First of all, the 'G-meter' typically shows the normal acceleration: that is, along the body Z axis. It is 'vertical' for the pilot, but not necessarily vertical for the Earth. This is the critical acceleration component, the greatest one (because the wing can produce by far the greatest force on an airplane), the one that is always specified in the flight manual and the one that you might exceed in normal manoeuvring. This is why the device shows it and not the 'total' combined acceleration.
Thus, the gravity component as measured by the 'normal' accelerometer will depend on the airplane attitude. In terms of G, the factor will be $cos(pitch) \cdot cos(roll)$. For the given example, if we assume the glideslope angle -3°, pitch will be $14-3 = 11°$, and with zero roll, the measured G will be $cos(11°) \approx 0.98$. Very small difference indeed.
However, in a more extreme example of a vertical dive, the accelerometer will show zero (0G).
But that's not all. If we want to be really pedantic and measure beyond 1%, we'll need to take into account that the apparent gravity changes depending on the location. Given that the accelerometer shows, essentially, weight of a calibrated mass, it will indicate different G even if the aircraft is standing on the ground.
I'm not talking about gravity anomalies, which are truly miniscule. The biggest contributor is Earth rotation, which adds some centripetal acceleration, the more of it the closer to the equator you get. At the equator, it reduces your weight by almost $0.4$% with respect to the pole.
A further $0.1$% can be accounted by the fact that Earth radius is greater at the equator than at the poles - by about 21 km. This is typically a bigger contributor than the aircraft's own altitude (esp. given that glideslopes are usually near the ground).
Altogether, if we calibrated at a pole and then make an approach near the equator in the stated conditions, we can expect almost $2.5$% reduction of the indicated G-force.
In truth, the calibration is usually done for middle latitudes (and digital AHRS may calibrate before takeoff on the actual location), so the difference will be even smaller. But anyway, the biggest contributor - attitude - is unaffected by these small changes. Yet, it is still small by itself, often below the accuracy of the G-meter.
the G meter measures acceleration. If you are descending at constant vertical speed or climbing at constant vertical speed, it will read 1G.
The difference in the force of gravity between that at the earth's surface and that at 35,000 feet is far too small to be detected by an aircraft's G meter.
If you fly straight ahead and at constant speed (vertical and horiz) then the only acceleration on the airframe will be the gravity, so total 1G as you normally understand it.
Wether the instrument will actually display 1G depends on calibration (funny trivia, gravity acc. at 35'000 feet is less than 1G since you are further from the Earth) and also depends if instrument shows total acceleration or longitudinal/vertical components.
To put this another way, just because you are descending it will not show <1G, just as it will not show >1G if you are climbing. If you think about it, it would violate a lot of Einstein's relativity postulates as you would be basically be able to tell how you are moving in relation to Earth by using a simple self contained accelerometer.