The answer is yes. Here is a rather well known metrication theorem:
Theorem: Let $a:\mathbb{N}\rightarrow\mathbb{Z}$ be an injective function. The sequence $(x a(n):n\in\mathbb{N})$ is uniformly distributed (u.d.) mod $1$ for Lebesgue almost all $x\in\mathbb{R}$.
A proof of this result can be found in the classical book Kuipers, L. and Niederreiter, H., Uniform Distribution of Sequences, Dover Publications Inc., New York, 2006 (reprint of the 1974 edition). pp. 32-33.
More general results (proved by Koksma) that include the situation of the OP are the following:
Theorem: Suppose $u_n:[a,b]\rightarrow\mathbb{R}$ is a sequence of functions in $\mathcal{C}^1[a,b]$ such that for any $n\neq m$, $u'_n-u'_n$ is monotone and
$$\|u'_m-u'_n\|_\infty\geq K$$
for some constant $K>0$. Then, for (Lebesgue) almost all $x\in[a,b]$, the sequence $(u_n(x))$ is uniformly distributed (u.d.) mod $1$.
This results yields the following well known results:
Corollary 1: Let $c>0$ and $F:\mathbb{N}\rightarrow(0,\infty)$ such that
$$|F(n)-F(m)|\geq c\delta_{nm}$$
For any $a\in\mathbb{R}\setminus\{0\}$,
the sequence $\big(ax^{F(n)}:n\in\mathbb{N}\big)$ is uniformly distributed mod $1$ for Lebesgue almost all $x\geq1$.
Corollary 2: For (Lebesgue) For $a\neq0$, the sequence $(ax^n:n\in\mathbb{N})$ is u.d. mod $1$ for (Lebesgue) almost all $x\geq 1$.
Corollary 3: Let $c>0$. If $(\lambda_n:n\in\mathbb{N})\subset\mathbb{R}$ is a sequence such that
$|\lambda_n-\lambda_m|\geq c\delta_{nm}$, then $(x\lambda_n:n\in\mathbb{N})$ is u.d. mod $1$ for (Lebesgue) almost all $x\in\mathbb{R}$.
The well known exceptions to Corollary 1 are the powers of Pisot-Vijayaraghavan numbers (PV-numbers). An algebraic number $\xi\in\mathbb{C}$ with $|\xi|>1$ is a P.V number if there is a monomial $p(z)=z^n+a_{n-1}z^{n-1}+\ldots + a_0$ with $a_j\in\mathbb{Z}$ such that $p(\xi)=0$, $p$ is irreducible over $\mathbb{Q}$ and all other roots $\xi'$ of $p$ satisfy $|\xi'|<1$.
Let $\{\;\}$ denote the fractional part function. It is known that
Theorem: If $\xi$ is a real PV-number, then the only limit points of the sequence $(\{\xi^n\}:n\in\mathbb{N})$ are $0$ and $1$.
See Kuipers (idem) pp. 34-36
