I recently asked a related question (inspired by this one!) . I accepted an answer but also provided an answer of my own. That answer illustrates the fact that for real $r$ one can arrange to have $\lfloor r^n \rfloor$ all even. One can pick the successive terms of $\lfloor r \rfloor, \lfloor r^2 \rfloor, \lfloor r^3 \rfloor,
\ldots, \lfloor r^n \rfloor, \ldots$ fairly freely. By this I mean being slightly vague (or uncommitted) about the value of $r$ and specifying it by providing the sequence of floor values one term at a time. It is easy to preserve freedom of choice enough to have about $\lfloor r \rfloor$ consistent choices at each stage (the smallest of which should be rejected in order to preserve the freedom for the following stage).
Let me relate it to other similar problems and implicitly suggest that for almost all non-integer rationals greater than $1$ there will be infinitely many prime floors. That we might not be able to prove it true for even one rational and that there may not be even one counter-example (at least that we can discover).
Given a positive real $r$ and a real "base" $b \gt 1$ define the base $b$ expansion of $r$ to be the integer sequence $r_b=(x_0,x_1,x_2,\cdots)$ with $x_k=\lfloor rb^k \rfloor$. So this particular integer sequence has $\lim\frac{x_{k+1}}{x_k}=b.$
Does the sequence $\pi_{10}=3,31,314,3141,\cdots$ contain infinitely many primes? One would expect so and even that there are about $\frac{\ln{N}}{\ln{10}}$ up to $n=N$ (in an appropriate sense), but one also does not expect to see a proof.
What about the same problem with $b=3$ or $b=4$? Same mutatis-mutandis.
what about $r_{10}=\lfloor 10^n r \rfloor$ for other reals $r?$ Well, we can pick the decimal expansion to avoid primes but otherwise we might predict infinitely many. The case $r=1/9$ is the question "Are there infinitely many prime repunits?" We know that we can only get a prime for $n$ prime, but might guess that something like $\frac{\ln{\ln{N}}}{\ln{10}}$ is the correct growth rate. We could switch this to questions about Mersenne primes by using the non-terminating binary expansion of $c=1$ (or $c=1/2$).
What about $\pi_{\pi}=\lfloor \pi^n \pi \rfloor$ in place of $\lfloor 3^n \pi \rfloor$? Either is a sequence of integers each roughly triple the one before and nothing much should be different.
The main point of my answer to my question is that we can build up a feasible integer sequence $\lfloor r^n \rfloor$ or $\lfloor c r^n \rfloor$ term by term somewhat like a decimal expansion. Eventually typically having about $r$ choices for the next term. At times the next choice can be forced by previous ones but that does not persist. So we can avoid primes if we build $r$ by startiing with the expansion. There are specific reals $r$ with special behavior (see the rest of this answer) but no reason to expect that any non-integer rationals are among them.
In special cases like $r=\sqrt[3]{7}$ we know that every third term of $r_r$ is non-prime, but that still suggests infinitely many prime floors, just at a slower growth rate.
There are certain algebraic integers (Pisot-Vijarayaghavan numbers) $r$ where the distance from $r^n$ to the nearest integer goes to $0$ exponentially fast. Then there is a recurrence relation for $\lfloor r^n \rfloor$ and things can sometimes be cooked to make these all even (for example). With the right definitions ($\lceil rb^n \rceil$ for $r=\frac{1}{\sqrt{5}}, b=\frac{1+\sqrt{5}}2$ ) we get the question "are there infinitely many (odd) prime Fibonacci numbers?" As with repunits, only prime $n$ could give a prime here and the situation should be about the same.
Suppose that we look at possible sequences $\lceil r^n \rceil.$ Then every initial segment of the sequence of Fermat numbers $3,5,9,17\cdots$ is feasible and the nested intervals are specifying a real in $\cap (2,\sqrt[k]{2^k+1}].$ These intervals have empty intersection but $\cap [2,\sqrt[k]{2^k+1}]=\{{2\}}$. In this non-standard expansion only the terms with $n=2^j$ could possibly be prime. Hueristics suggest that we know all the Fermat Primes (just $0 \le j \le 4$) but altering the heuristic arguments can (evidently) support a case that a positive proportion of choices $n=2^j$ give Fermat primes.
In conclusion: Give a real constant $r$ and a base $b \in \mathbf{R}$, the integer sequence $\lfloor r b^n\rfloor$ gives the base $b$ expansion of $r$. This is a sequence of integers each roughly $b$ times the previous and generally the previous choices give an interval of length $b$ for possible values of the next term. If we start with the expansion we can avoid primes, but a "random" constant $r\ $ should yield a predictable infinite number of primes subject to adjustments we feel that we understand.
Not that much is different if we use $b=r$. There are nameable reals with exceptional expansions $r_r$ (Integers and Pisot numbers), $r=\sqrt[j]{k}$ a root of an integer does certain special things when $j \mid n$. We have a lot of latitude to construct $r_r$, the "base $r$" expansion of $r$ to get an integer sequence enjoying certain properties, but there is no reason to think that this would lead to a rational number.
