I don't know much about the operation of a Telsa Turbine but I can talk in general about what affects the stiffness of an object.
Short answer
The geometry of the disc and the material of the disc both affect the stiffness of the disc. Smaller radius = stiffer. Thicker disc = stiffer. If you are looking for a trade-off between stiffness and weight then you can employ tricks like ribs or hollow pockets inside the disc. The material property you are looking for is the Young's modulus, $E$. Higher $E$ = stiffer material. You can find a list of Young's Moduli for various materials here:http://www.engineeringtoolbox.com/young-modulus-d_417.html. Modern HDD's are made glass/ceramic composites apparently. If I had to suggest a material and money was no problem, I would say graphene or some other nano-engineered material where you could design the stiffness properties at the molecular level. In terms of affordable materials however, you probably won't be able to find anything better than the HDD platter. Of course you will also need to worry about corrosion resistance, temperature stability, etc. due to the environment in which you are operating.
Now the long answer
Stiffness is generally characterized in engineering as the relationship between the force applied to an object and the amount that object deflects:
$$\delta = f(F,K)$$
Where $\delta$ is the displacement of the point where the force is applied, $F$ is an applied force and $K$ is the stiffness. In the equation above, $K$ is not necessarily a constant with respect to time or deflection. In fact, it can be a function of many things, including but not limited to the deflection, the rate of deflection, the rate of load application, the temperature of the material, etc. The relationship between force and displacement is often simplified to a linear relationship, where $K$ is assumed to be constant. In this case, K is often called the spring constant.
$$F = K\delta$$
If we ignore those more complicated effects, then at its most basic level the stiffness of an object is determined by two things: the material properties and the geometry.
The material property most relevant to stiffness is the Young's modulus, $E$. Higher Young's Modulus = stiffer material. From what I found (https://en.wikipedia.org/wiki/Hard_disk_drive_platter) modern HDD platters are made of a glass/ceramic composite, which I'm guessing would be a proprietary, specially-engineered material for each manufacturer. The Young's modulus for these materials is probably not easily available online. As a comparison however, glass has a Young's modulus of 50-90 GPa. Aluminum is what they used to use for HDD platters, which has a Young's modulus of around 70 GPa. Stainless steel has a Young's modulus of around 180 GPa. If money wasn't an issue then you could make a specially engineered carbon-based structure, like graphene which has a Young's modulus of around 1000 GPa. At the end of the day, you're going to be hard pressed to find a material that is as thin as a HDD platter but stiffer. I don't have a source for this, but in my experience steel can't be machined that thinly (it would get seriously warped due to the heat of the machining process), so you'd need to use other techniques to produce the disc. That's why these things are made with glass and ceramics, which can be assembled with chemical techniques instead of mechanical ones.
The geometry part is where things get really complicated.
Here is a diagram of how I am picturing your situation:
There is a circular disk pinned in the center by some kind of axle. The plate deflects to a new position (dotted line) when a force is applied. If I understand correctly, the stiffness that you are worried about is the "floppy" direction of the floppy disc, i.e. the bending stiffness, as I have drawn it and not the torsional stiffness or compression stiffness.
This is a case of plate bending and here is a Wikipedia article where you can read about how that works. https://en.wikipedia.org/wiki/Bending_of_plates. Yes, the math is very nasty.
To get a better idea of how geometry affects stiffness, I propose a simplified case of cantilever beam bending:
There is a known solution for this case (source: https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory).
$$F = \frac{3EI}{L^3}\delta$$
So where does the "thickness", $h$, come in? $I$ is called the area moment of inertia and it is calculated from the cross-section of the beam. In the simple rectangular beam case that I have drawn, $I$ is equal to the following:
$$I = \frac{bh^3}{12}$$
Check out my source here: https://en.wikipedia.org/wiki/List_of_second_moments_of_area. That page contains many different cross section shapes. Some are more efficient than others. A cross section that is often used in the construction industry is the I-beam (https://en.wikipedia.org/wiki/I-beam) because it handles bending and shear loads quite well, while also conserving material.
So what can we conclude from the simpler case of cantilever beam bending for our plate bending problem? If you have a solid cross-section for a flat disc, the only ways to improve the stiffness geometrically is to reduce the radius of the disc (analogous to $L$) or increase its thickness (analogous to $h$). If you aren't constrained to solid cross-sections or flat discs then you can use ribs or pockets in your disc to reduce material usage while maintaining stiffness. Again, this is ignoring the more complicated effects that can occur with plate bending.
Maybe somebody else can look into whether or not reducing the radius of the disc will affect the functionality of the turbine. Fluid dynamics isn't my specialty!
Wow. Bit of an essay there. Hope this helps.
