Suppose the input of SHA-512 is 512 bits of data (so exactly the same size as the output). If I would test every possible combination (so $2^{512}$ calculations), then will the output also be exactly $2^{512}$ different hashes, or will collisions occur?
It would be astonishing if SHA-512 turned out to be a permutation on 512-bit inputs, so no, absent a publication-worthy breakthrough in SHA-512 analysis, the output would not be exactly $2^{512}$ distinct hashes.
If collisions occur, would the amount of collisions and the 'size' of the collisions (approximately) be the same as statistics would predict after randomly generating $2^{512}$ 512-bit strings? (With 'size' I mean the amount of times a specific hash occurs)
As far as we know, yes. The usual model is a uniform random function. Under each distinct input, the output is an independent uniform random bit string. This model is like writing down a giant book of outputs with $2^{512}$ pages, one page per input having the corresponding 512-bit output, where every output is chosen independently and uniformly at random—which is exactly what you just described.
If we examine the structure of SHA-512 under the hood—a block cipher, which we might call ‘SHACAL-512’ (related to SHACAL-2, which is the block cipher underlying SHA-256), in Davies–Meyer composition—for 512-bit message $m$, we have $H(m) = E_{\operatorname{pad}(m)}(h_0)$ where $h_0$ is the standard initialization vector. In the ideal cipher model, every distinct 1024-bit string $\operatorname{pad}(m)$ gives rise to an independent uniform random permutation $E_{\operatorname{pad}}(m)$, whose output on a fixed input string $h_0$ is then an independent uniform random 512-bit string. Absent cryptanalysis of ‘SHACAL-512’, there isn't a much better model for this either.
