It seems to me that at least for $k_0=0$ these sums always satisfy a linear recurrence. Maybe one can guess the general form of this recurrence. For example (using FriCAS):
(1) -> )expose RECOP
(1) -> floorsum(p, q, k, n) == reduce(+, [floor(i*p/q)^k for i in 0..n])
(2) -> guessEq(p, q, k) == (N := 50; while empty?(e := guessPRec([floorsum(13, 17, k, n) for n in 0..N], safety==100, maxDegree==0)) repeat N := N+1; getEq first e = 0)
(3) -> guessEq(13,17,0)
(3) - f(n + 1) + f(n) + 1= 0
(4) -> guessEq(13,17,1)
(4) - f(n + 18) + f(n + 17) + f(n + 1) - f(n) + 13= 0
(5) -> guessEq(13,17,2)
(5)
- f(n + 35) + f(n + 34) + 2f(n + 18) - 2f(n + 17) - f(n + 1) + f(n) + 338= 0
(6) -> guessEq(13,17,3)
(6)
- f(n + 52) + f(n + 51) + 3f(n + 35) - 3f(n + 34) - 3f(n + 18) + 3f(n + 17)
+
f(n + 1) - f(n) + 13182
=
0
(7) -> guessEq(13,17,4)
(7)
- f(n + 69) + f(n + 68) + 4f(n + 52) - 4f(n + 51) - 6f(n + 35) + 6f(n + 34)
+
4f(n + 18) - 4f(n + 17) - f(n + 1) + f(n) + 685464
=
0
(9) -> guessEq(13,17,5)
(9)
- f(n + 86) + f(n + 85) + 5f(n + 69) - 5f(n + 68) - 10f(n + 52)
+
10f(n + 51) + 10f(n + 35) - 10f(n + 34) - 5f(n + 18) + 5f(n + 17)
+
f(n + 1) - f(n) + 44555160
=
0
(12) -> guessPRec [1,13,338,13182,685464,44555160]
(12) [[f(n): - f(n + 1) + (13n + 13)f(n)= 0,f(0)= 1]]
(13) -> guessPRec [1,18,35,52,69,86]
(13) [17n + 1]
It is useful to classify arithmetic functions $f(n)$ involving the primes into two classes: those whose Dirichlet series involves the zeta function (or something similar, such as a Dirichlet L-function) in the numerator, and those that involve zeta functions in the denominator. Examples of the former type include the constant function 1 and the various divisor sum functions. Examples of the latter type include the Mobius and von Mangoldt functions. The functions of the latter type are much harder to control, because by their nature they are very sensitive to the location of zeroes of the zeta function, and in particular are sensitive to the truth of statements such as the Riemann hypothesis. Functions of the former type, by contrast, pay almost no attention to the location of these zeroes.
Because of this difference in sensitivity to zeta zeroes, it is highly unlikely that one can rely solely on asymptotics for sums of functions of the first type to conclude useful asymptotics on functions of the second type, and the series expansions which connect the two are likely to have very poor convergence properties (much as Taylor series such as $\log(1-x) = -x - x^2/2 - x^3/3 - \ldots$ must necessarily have poor convergence properties near the singularity x=1).
On the other hand, there are certainly useful hyperbola-type identities for sums involving functions of the second type, such as Vaughan's identity. The catch is that the sums that come out of such identities, if they are to be useful, must still involve at least one function of the second type, since one cannot hope for the influence of zeta zeroes to magically disappear (unless the identity is so badly convergent that one can hide this influence inside the accumulated error terms in the asymptotics; but in such cases, the identity is likely to be useless for the purpose of producing asymptotics).
It is also possible to use methods of sieve theory to upper bound functions of the second type, such as $\Lambda$, by divisor sums which are basically of the first type, and indeed such upper bound sieves can be very useful in analytic number theory. However, they do not seem capable, by themselves, of producing asymptotics for functions of the second type (although with the addition of non-trivial ingredients outside of sieve theory, this is occasionally possible, e.g. with the Friedlander-Iwaniec theorem).
Best Answer
Let me try to provide a partial answer in the case when $0<\beta<2$ much in line with what Terry suggested in the comments. Perhaps it is possible to extend this method to all $\beta>2$ but the computations get more complicated.
Using Taylor expansion and $\lfloor n^\theta\rfloor=n^\theta-\{n^\theta\}$ we obtain $$ \lfloor n^\theta\rfloor^\beta = n^{\theta\beta}+\mathrm{o}_{n\to\infty}(1) $$ when $0<\beta<1$ and $$ \lfloor n^\theta\rfloor^\beta = n^{\theta\beta}-\beta \{n^\theta\}n^{\theta(\beta-1)}+\mathrm{o}_{n\to\infty}(1) $$ when $1<\beta<2$. In the first case, one has $\sum_{1\leq n\leq N}e(\lfloor n^\theta\rfloor^\beta)\approx\sum_{1\leq n\leq N}e(n^{\theta\beta})$ and hence cancellation occurs exactly when $\theta\beta$ is not an integer (one can use the Kusmin-Landau inequality/ van der Corput method to obtain good bounds on the cancellation in this case, see for example the book by Graham and Kolesnik: https://doi.org/10.1017/CBO9780511661976).
So lets focus on the second case when $1<\beta<2$, which seems more complicated. Using the notation from the comments, let $$ H(\mathbb{R})=\begin{pmatrix} 1 & \mathbb{R} & \mathbb{R} \\ 0 & 1 & \mathbb{R} \\ 0 & 0 & 1 \end{pmatrix}~~~\text{and}~~~H(\mathbb{Z})=\begin{pmatrix} 1 & \mathbb{Z} & \mathbb{Z} \\ 0 & 1 & \mathbb{Z} \\ 0 & 0 & 1 \end{pmatrix}. $$ Consider $G=\mathbb{R}\times H(\mathbb{R})$ and $\Gamma=\mathbb{Z}\times H(\mathbb{Z})$ and take their quotient to obtain the nilmanifold $G/\Gamma=\mathbb{R}/\mathbb{Z}\times H(\mathbb{R})/H(\mathbb{Z})$. Consider the function $$ F\left(u,\begin{pmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{pmatrix}\right)= e(u-z+x\lfloor y\rfloor), $$ which is a Riemann integrable and well defined function on $X$. (This is because $(\{x\},\{y\}, \{z-x\lfloor y\rfloor\})$ is a fundamental domain for the Heisenberg nilmanifold, see here for details: https://en.wikipedia.org/wiki/Nilmanifold). Now consider the sequence $$ g(n)=\left(n^{\theta\beta},\begin{pmatrix} 1 & \beta n^{\theta(\beta-1)} & \beta n^{\theta\beta} \\ 0 & 1 & n^\theta \\ 0 & 0 & 1 \end{pmatrix}\right) $$ Note that with this choice of $X$, $F$,and $g$ we have $$ \frac{1}{N}\sum_{1\leq n\leq N}e(\lfloor n^\theta\rfloor^\beta) \approx \frac{1}{N}\sum_{1\leq n\leq N}e(n^{\theta\beta}-\beta \{n^\theta\}n^{\theta(\beta-1)})=\frac{1}{N}\sum_{1\leq n\leq N} F(g(n)). $$ Due to the main result in https://arxiv.org/abs/2006.02028 we know that the sequence $g(n)$ is uniformly distributed in the nilmanifold $X$ if and only if $(n^{\theta\beta},\beta n^{\theta(\beta-1)},n^\theta)$ is uniformly distributed in $\mathbb{T}^3=\mathbb{R}^3/\mathbb{Z}^3$. So if neither $\theta\beta$ nor $\theta(\beta-1)$ are integers then the sequence $g(n)$ is uniformly distributed in $X$ and hence $$ \frac{1}{N}\sum_{1\leq n\leq N} F(g(n))\rightarrow \int F\,d\mu_X, $$ where $\mu_X$ is the natural (and normalized) Haar measure on $X$. But since $\int F\,d\mu_X=0$, it follows that $$ \frac{1}{N}\sum_{1\leq n\leq N}e(\lfloor n^\theta\rfloor^\beta)\rightarrow 0. $$
Note that if $\theta(\beta-1)$ is an integer then the sequence $g(n)$ is not uniformly distributed on the entire nilmanifold $X$ but rather distributes uniformly on a sub-nilmanifold $Y\subset X$. But due to the simple nature of $F$, it is not so hard to see that for any such sub-nilmanifold $Y$ one has $\int F\,d\mu_Y=0$, which in turn still implies cancellation. So the only condition seems to be that $\theta\beta$ is not an integer.
Please let me know if this is helpful or if you have any questions.