In general, no. For the double integral $\iint_T f(x,y)\,dx\,dy$ will be the height on any side, as is seen by turning the triangle with one side parallel to an axis and performing the integral. So at least, $T$ has to be equilateral. I don't know the answer in that case.

**Edit:** Wait, wait – the same trick works even if I turn the triangle at any angle, hence all heights (defined as $\sup_{RT} y-\inf_{RT} y$ where $R$ is a rotation) must be the same. That is never true for a triangle, and limits the number of convex sets seriously – but there are still non-circles that might satisfy the criterion. Once more, I don't know the answer, but for circles at least, it should in principle be straightforward to check if a radially symmetric function will do. And of course, if there is a solution, there is a radially symmetric one, as can be seen by rotating the solution and taking the average of all its rotated variants.

**Edit**^{2}: For the unit disk (and radially symmetric $f$) the answer should in principle be obtainable by the Abel transform (which is really nothing but the Radon transform on radially symmetric functions). The required Abel transform $\Phi$ should be the characteristic function of the interval $[0,1]$ (we only use positive $x$ due to symmetry), and the inverse Abel transform provides the answer:
$$f(r)=\frac{-1}{\pi}\int_r^\infty \frac{\Phi'(x)\,dx}{\sqrt{x^2-r^2}}=\frac{1}{\pi\sqrt{1-r^2}}$$
when $0\le r<1$. Being lazy, I checked the answer using Maple, and it seems right.

**Addendum:** Anton Petrunin pointed out in a comment that the above measure is the push-forward of the surface measure on the unit sphere on the unit disk under projection. It is well-known that the surface area of the portion of a disk between two parallel planes depends only on the distance between the planes (and is proportional to said distance), which ties in nicely with the desired property of $f$ on the disk.