Assume that the point mass, $m$ has two tiny thrusters, mounted so as to exert purely tangential force in the plane of the circular motion, one clockwise, and the other counter-clockwise.
The magnitude of the constant velocity of the mass is $v$, and the radius of the circle is $r$.
Measure the position of the point mass in the standard Cartesian coordinate way: angles are measured from the positive X-axis, counter-clockwise positive.
At the point where the mass is at a position angle $\theta$. the total radial force inward on the mass, $F_R$ is given by the centripetal force equation:$$F_R=\frac{mv^2}{r}$$
There are two forces that supply this radial force: the tension, $T$ in the string, and the inward radial component of the force of gravity:$$F_{G-R}=mg\sin(\theta)$$So:$$\frac{mv^2}{r}=T+mg\sin(\theta)$$and:$$T=\frac{mv^2}{r}-mg\sin(\theta)$$Note that this implies that:$$v >=\sqrt{rg}$$ or the string tension will become negative near the top of the circle, an impossibility.
The conditions of the question also require that at all times the net tangential force, $F_T$, be zero. The tangential component of the force of gravity, $F_{G-T}$ is given by:$$F_{G-T}=-mg\cos(\theta)$$where a positive force implies counter-clockwise force. The thrusters are needed to supply the exact opposite force to the mass at all times.
The proper question should be "Can a varying centripetal force produce circular motion?" The answer is yes.
The (much-abused) expression $$m\frac{v^2}{r}$$ tells us what the instantaneous transverse (perpendicular) component of force must be in order for an object of mass $m$ must experience if it is travelling at instantaneous speed $v$ and is turning around an instantaneous center of curvature a distance $r$ away. It is not a force itself. It has force units, but it is the product of a mass times an acceleration. There must be some real forces which combine to result in the same net value as the product $m \times \frac{v^2}{r}$.
This is true every time the velocity of a particle changes direction.
If we apply it to circular motion, then $r$ is the radius of a circle. If we apply it to elliptical or parabolic or hyperbolic motion, then $r$ is constantly changing. If we apply it to straight line motion, then $r\rightarrow \infty$.
Back to circular motion, we can now write $${\large\Sigma}F_r=-m\frac{v^2}{r}.$$
This must be true instantaneously during every moment because the velocity direction is always changing. (The minus sign shows that the net force must be toward the center. Positive radial direction in polar coordinates is traditionally away from the center. This is important when considering a mixture of forces, some of which might, by themselves, pull the object away from the center.)
For uniform circular motion, the speed is constant, so the magnitude of the radial component of the net force must be constant. Notice that we could have two or more changing forces which still produce uniform circular motion as long as the net tangential component is zero (no speed change) and the net radial component adds up to $mv^2/r$ toward the center.
For non-uniform circular motion, we still have a constant $r$, but the speed is changing. That means that the radial component of the net force must change. How can this happen? Consider a mass attached to a string, swinging in vertical circle with non-constant speed. The mass moves fast at the bottom and slows down, but stays on the circular path at the top. The gravitational pull on the mass is constant from the Earth-bound reference frame. But it is not always a radial force. That means that the tension magnitude in the string must be changing in such a way that the net radial component always has a magnitude toward the center of $mv^2/r$.
If we define $\theta$ to be the angle measured from a vertical line passing through the center, with $\theta=0$ being at the bottom of the circle, then
$$mg\cos\theta-F_\text{ten}=-m\frac{v^2}{r}$$
$$F_\text{ten}=m\left(g\cos\theta+\frac{v_{\theta}^2}{r}\right)$$
where $v_{\theta}$ is the speed at angle position $\theta$ (not the angular speed).
We also know that the tension force will not affect the speed (it acts perpendicularly to the velocity), but the gravity force will change the speed like $$\frac{1}{2}mv_\text{bottom}^2=\frac{1}{2}mv_{\theta}^2 + mgr(1-\cos\theta).$$
From there it's an algebraic exercise to show that the tension force must change, and that there is a minimum required speed at the bottom for the mass to stay on the vertical circle.
Summary: for non-uniform circular motion, the radial magnitude of the net force must change, and it can.
Best Answer
The biggest problem will be making any practical measurement with the vertical plane rotation. The tension in the string will no longer be constant, as you clearly understand, but that means that the hanging slotted weight (m) will never balance the tension (well, actually it will 'balance' at two specific angles, bjut that is like saying that a stopped clock is correct twice a day).
So during each 2Pi rotation the tension will either be greater than mg, so the weight will accelerate upwards, or will be less than mg, and the weight will accelerate downwards. Measuring the motion of a weight where it's acceleration is varying continuously will be difficult if at all practical. In fact it isn't clear to me that you will have any sort of equilibrium (as you do in the horizontal experiment) - I believe that it is likely that at any constant rotation rate the hanging mass will either drop to the floor or will accelerate upwards until it hits the 'blue' box in your diagram.
In practice, it might be better to replace the hanging mass with a spring balance where, with care you could at least measure the maximum tension in the string as a function of rotational velocity and radius of rotation. Alternately, if you want to keep the hanging mass (or if that is all you have to use) one nice experiment is to drop the rotating mass from various heights (and with varying string length) to see at combinations of drop height and arc radius provide sufficient tension to 'just' lift the slotted weight as the rotating mass reaches the bottom of its arc. (see sketch below). Of course that makes it a very different experiment to the horizontal rotating mass experiment.
Just calculating the dynamics of the vertical rotating system (with hanging mass) might have some fairly challenging maths involved. But I think you have the right research spirit to go a long way at whatever you try.