If you look for the integrals to be defined in the sens of Lebesgue, the answer is no. Indeed, if there is an $n_0$ such that $\int_0^1 |r^{2n_0 + 1}\omega(r)| \, dr < +\infty$, then by dominated convergence, $\int_0^1 r^{2n + 1}\omega(r) \, dr \rightarrow 0$.
If you look for an $\omega$ such that the $\int_0^1 r^{2n + 1}\omega(r) \, dr$ are defined as unproper integrals at zero for example with $\int_0^1 r^{2n + 1}\omega(r) \, dr := \lim_{\varepsilon \rightarrow 0} \int_\varepsilon^1 r^{2n + 1}\omega(r) \, dr$, it might be possible (but I am not sure). For this, consider the following lemma :
For all interval $I \subset \mathbb{R}$ of non empty interior, for all integer $n$, there exists a real polynomial odd $P$ of degree $\leqslant 2n + 1$ such that $\int_I r^{2n + 1}P(r) dr = 1$ and $\int_I r^{2m + 1}P(r) dr = 0$ when $m < n$.
To prove this, let $E = \mathrm{Span}(r \mapsto r^{2m + 1},m \leqslant n)$ which is a finite dimensional vector space, endowed with the scalar product $\left<f,g\right> = \int_I f(r)g(r) \, dr$. The family $b_m : r \mapsto r^{2m + 1}$ is free hence it is a basis of $E$. Therefore, we can define the linear form $\mu : \sum_{m = 0}^n a_mb_m \mapsto a_n$ which verifies $\mu(b_n) = 1$ and $\mu(b_m) = 0$ if $m < n$. By Riesz' representation theorem, there exists $P \in E$ (thus $P$ is a real odd polynomial of degree $\leqslant n$) such that for all $f$, $\mu(f) = \int_I f(r)P(r) \, dr$. $P$ fits.
Now that we have this lemma, we will built recursively a sequence $(\omega)_{n \geqslant -1}$ of continuous by parts functions of $[0,1]$ such that, for all $n$,
$(1)$ $\omega_n = 0$ on $\left[0,\frac{1}{n + 2}\right[$,
$(2)$ $\int_0^1 r^{2m + 1}\omega_n(r) \, dr = m$ if $m \leqslant n$,
$(3)$ $\omega_n = \omega_{n - 1}$ on $\left[\frac{1}{n + 1},1\right]$ if $n \geqslant 0$.
First of all, set $\omega_{-1} = 0$ on $[0,1]$, which obviously satisfies the conditions. Now, take $n \geqslant 0$ and assume that $\omega_{n - 1}$ is built. Let $I = \left[\frac{1}{n + 2},\frac{1}{n + 1}\right]$ and let $P$ be a polynomial such that $\int_I r^{2m + 1}P(r) \, dr = \delta_{n,m}$ if $m \leqslant n$, given by the lemma. Let,
$$
\omega_n : \left\{\begin{array}{rcl} [0,1] & \rightarrow & \mathbb{R} \\ 0 \leqslant r < \frac{1}{n + 2} & \mapsto & 0 \\ \frac{1}{n + 2} \leqslant r < \frac{1}{n + 1} & \mapsto & \displaystyle \left(n - \int_0^1 t^{2n + 1}\omega_{n - 1}(t) \, dt\right)P(r) \\ \frac{1}{n + 1} \leqslant r \leqslant 1 & \mapsto & \omega_{n - 1}(r) \end{array}\right.
$$
Conditions $(1)$ and $(3)$ are immediately satisfied and for condition $(2)$, let $c = n - \int_0^1 t^{2n + 1}\omega_{n - 1}(t) \, dt$. We have for all $m \leqslant n - 1$,
\begin{align*}
\int_0^1 r^{2m + 1}\omega_n(r) \, dr & = \int_0^{1/(n + 2)} 0 \, dr + \int_{1/(n + 2)}^{1/(n + 1)} cr^{2m + 1}P(r) \, dr + \int_{1/(n + 1)}^1 r^{2m + 1}\omega_{n - 1}(r) \, dr\\
& = c\int_I r^{2m + 1}P(r) \, dr + \int_0^1 r^{2n + 1}\omega_{n - 1}(r) \, dr\\
& = m.
\end{align*}
And if $m = n$,
\begin{align*}
\int_0^1 r^{2n + 1}\omega_n(r) \, dr & = c\int_I r^{2n + 1}P(r) \, dr + \int_0^1 r^{2m + 1}\omega_{n - 1}(r) \, dr\\
& = c + \int_0^1 r^{2m + 1}\omega_{n - 1}(r) \, dr\\
& = n.
\end{align*}
Now, let $\omega(0) = 0$ and when $r > 0$, there exists an integer $n$ such that $\frac{1}{n + 2} \leqslant r$. In this case, let $\omega(r) = \omega_n(r)$ (it deosn't depend on the choice of $n$ because of condition $(3)$). By the way, we can easily check that we have $\omega_n \rightarrow \omega$ simply. In particular, it implies that $\omega$ is measurable and you can easily verify that for all $n$,
$$
\int_{1/m}^1 r^{2n + 1}\omega(r) \, dr \rightarrow n
$$
when $m$ grows towards infinity and is an integer. However, it absolutely doesn't mean that $\int_\varepsilon^1 r^{2n + 1}\omega(r) \, dr \rightarrow n$ when $\varepsilon \rightarrow 0$. Maybe you can try to bound the absolute value of the $P$ given by the lemma to prove that the $\int_\varepsilon^{1/\lfloor 1/\varepsilon \rfloor} r^{2n + 1}\omega(r) \, dr$ converge toward zero, but I don't garantee it is possible.
