We claim that $a$ can be any rational number such that $\frac{1}{a}$ is not an integer. If $a$ is replaced by $-a$, then $x$ can be replaced by $\frac{1}{x}$. WLOG let $a \ge 0$. If $a = \frac{1}{c}$ for some integer $c$ and $q = \frac{m}{n}$ for some relatively prime positive integers $m$ and $n$, then if $x = \left(m + \frac{1}{n}\right)^c$, $\lfloor x^a \rfloor \{x^a\} = m \cdot \frac{1}{n} = q$. Thus, the equation has a rational solution for all positive rationals $q$. If $a = 0$, then $q$ can be any nonzero rational number. If $a = \frac{b}{c}$ for relatively prime positive integers $b \ge 2$ and $c$, then we claim that the equation has no rational solutions when $q = \frac{1}{n^b}$, where $n$ is any positive integer. First, note that since $x^a \ge \lfloor x^a \rfloor = \frac{q}{\{x^a\}} > 0$, then $x > 0$. Then, we see that $x^\frac{b}{c} = x^a = \lfloor x^a \rfloor + \{x^a\} = \lfloor x^a \rfloor + \frac{q}{\lfloor x^a \rfloor} \in \mathbb{Q}$, therefore, $x = y^c$ for some rational $y$. So $\lfloor y^b \rfloor \{y^b\} = q$. Let $k = \lfloor y^b \rfloor$, then $y^b = k + \frac{q}{k} = \frac{n^bk^2 + 1}{n^bk}$. Since $gcd(n^bk^2 + 1, n^bk) = gcd(1, n^bk) = 1$, $k = u^b$ and $n^bk^2 + 1 = v^b$ for some positive integers $u$ and $v$. Thus, $v^b - (nu^2)^b = 1$, and since $b \ge 2$, by Mihailescu’s theorem, this has no solution in positive integers. Hence, there are infinitely many positive rationals $q$ such that the equation has no rational solutions $x$.

The problem is equivalent to find all $a\in\mathbb{Q}$ such that $f_a(x)=\lfloor x^a \rfloor \{x^a\}$ doesn't cover infinitely many points of $\mathbb{Q^+}$. I will prove that the answer is $\mathbb{Q}\setminus \{\frac{1}{n} | n\in\mathbb{Z}_ {\ne0}\}$. First of all, since $\{x^a\}\in \mathbb{Q}$, we get $x^a\in \mathbb{Q}$. Then if we check whether $\frac{m}{k}$ is a solution where $(m,k)=1$, since $x^{\frac{m}{k}}$ is rational, we get $x=x_1^{k}$ for some rational $k$ and it means that it's same with checking whether $m$ is a solution. Additionally, since $f_a(x)=f_{-a}(\frac{1}{x})$, it is it's enough to solve problem for $a\in \mathbb{N}_0$. Obviously, $a=0$ is a solution. For $a=1$ take $x=a+\frac{1}{b}$ to get any rational number $\frac{a}{b}$. So, $n=1$ doesn't work. That's why, $\frac{1}{z}$ is not solution for all $z\in\mathbb{Z}_{\ne 0}$. Now, let's prove that $a=n$ where $n\in\mathbb{N}_{>1}$ works. To prove this, I will show that $f_n(x)=\frac{1}{t^n}$ has no solution, where $t$ is a natural number. Obviously, if $f_n(-x)$ is solution, then so is $f_n(x)$, which means we can take $x>0$. Let $x=\frac{a}{b}$ and $a^n=b^n\cdot c+r$, where $\gcd(a,b)=\gcd(b,r)=1$ and $r<b^n$. Then we should solve $f_n(x)=\frac{cr}{b^n}=\frac{1}{t^n}$. Since $r\mid b^n$ and $\gcd(r,b)=1$, we get $r=1$. We have $t^n\mid b^n$, which means $b=ts$ for some natural $s$ and it gives that $c=s^n$. Now we have that $a^n=(bs)^n+1$. From Fermat's Last Theorem, we get it has no solution for $n\geq 3$ and when $n=2$, we get $(a-bs)\cdot (a+bs)=1$ which has a solution $b=0, a=1$ which is impossible since $a,b\in \mathbb{N}$. So, all $a\in\mathbb{N}_{>1}$ and $a=0$ are solutions which means all $a=\frac{n}{z}$ are solutions, where $n\in\mathbb{N}_{>1}\cup\{0\}$ and $z\in \mathbb{Z}_{\ne 0}$.