Is there a closed-form expression for $\sum_{k=1}^{n}\lfloor k^{q} \rfloor$ for $q \in \mathbb{Q}_{> 0}$?

It's unlikely that there is a closed form expression for what you want. Consider the expression $$ \sum_{k=1}^{n}\left\lfloor f(k)\right\rfloor, $$ where $f(k)$ is a monotonically increasing function with $f(1)=1$. Let $g(m)$ be the smallest integer value of $k$ such that $f(k)\ge m$; i.e., $g(m)=\left\lceil f^{-1}(m)\right\rceil$. Then the terms in the sum are at least as large as $1$ starting at $k = g(1)$, and at least as large as $2$ starting at $k= g(2)$, and so on. As long as $g(m)\le n$, each term contributes $n-g(m)+1$ to the sum. So $$ \sum_{k=1}^{n}\left\lfloor f(k)\right\rfloor = \sum_{m=1}^{\lfloor f(n)\rfloor} \left(n + 1 - \left\lceil f^{-1}(m)\right\rceil\right)=(n+1)\lfloor f(n)\rfloor - \sum_{m=1}^{\lfloor f(n)\rfloor} \left\lceil f^{-1}(m)\right\rceil. $$ This will simplify in certain cases where the inverse is integer-valued (so the ceiling function goes away) and "nice" (so the resulting sum can be evaluated). Two standard examples are when the inverse is a polynomial (e.g., $f(k)=k^{1/q}$, so $f^{-1}(m)=m^q$) and when the inverse is an exponential function (e.g., $f(k)=1+\log_b k$, so $f^{-1}(m)=b^{m-1}$). In your case you have $$ \sum_{k=1}^{n}\left\lfloor k^{2/3} \right\rfloor=(n+1)\left\lfloor n^{2/3}\right\rfloor - \sum_{m=1}^{\lfloor n^{2/3}\rfloor} \left\lceil m^{3/2}\right\rceil, $$ but this sum doesn't seem any easier to evaluate.