I think the uniform distribution mod1 of $\{p/\alpha\}$ is due to Vinogradov, and the asmptotic for primes in a Beatty sequence $\sim \frac{\pi(x)}{\alpha}$ is an immediate consequence. Indeed for $p$ to be equal to some $\lfloor k\alpha\rfloor$ it is equivalent to $1-\frac{1}{\alpha}<\frac{p}{\alpha}-\lfloor \frac{p}{\alpha}\rfloor<1$. So you just need the fractional part of $p/\alpha$ to be on a fixed interval of length $\alpha$ mod1.
On a related note this paper discusses the general sequence $q\lfloor \alpha n+\beta\rfloor +a$.
