Let $f:\mathbb{R}\to(0,\infty)$ a locally integrable function. I want to compare these two conditions $$\limsup_{r\to + \infty}\frac{r}{\int_{-r}^r f(x)dx}<+\infty. \tag{1}\label{1}$$ and $$\limsup_{r\to + \infty}\frac{r^{\frac{1}{p}}\left(\int_{-r}^r f(x)^qdx\right)^\frac{1}{q}}{\int_{-r}^r f(x)dx}<+\infty, \tag{2}\label{2}$$ with $p,q>1$ and $\frac{1}{p}+\frac{1}{q}=1.$
We can see that if $f$ is bounded then $\eqref{1}$ is stronger than $\eqref{2}$, because we would have $$r^{\frac{1}{p}}\left(\int_{-r}^r f(x)^qdx\right)^\frac{1}{q}\leq M 2^\frac{1}{q}r.$$ Now if $\inf_{s\in\mathbb{R}}f(s)=c>0$, we have $\eqref{2}$ stronger than $\eqref{1}$. Because $$c\ 2^\frac{1}{q} \ r\leq r^{\frac{1}{p}}\left(\int_{-r}^r f(x)^qdx\right)^\frac{1}{q}.$$ Of course if we have $c\leq f(s)\leq M$ with $c>0$, then $\eqref{1}$ and $\eqref{2}$ are equivalent.
I found that for example $f(x)=e^x$, $f$ satisfies $\eqref{1}$ but not $\eqref{2}$. I don't know if I can find a function $f$ which satisfies $\eqref{2}$ but not $\eqref{1}$. To search it, $f$ must necessarily satisfy $\inf_{s\in\mathbb{R}}f(s)=0$.
Provided $-\frac{1}{q}<\alpha<0$, the function $f(x)=x^\alpha \cdot 1_{(1,\infty)}$ does what you want.
Indeed, since $\alpha +1>0$ and $\alpha<0$, $$\frac{r}{\int_{-r}^rf(x)dx}=(\alpha+1)\frac{r}{r^{\alpha+1}-1}\sim \frac{\alpha + 1}{r^\alpha}\longrightarrow +\infty $$
and since $\alpha q+1>0$,$$ \frac{r^\frac{1}{p} \left( \int_{-r}^r f(x)^qdx \right)^\frac{1}{q}}{\int_{-r}^rf(x)dx}= \frac{\alpha +1}{(\alpha q+1)^\frac{1}{q}}\frac{r^\frac{1}{p}\left(r^{\alpha q+1}-1\right)^\frac{1}{q}}{r^{\alpha+1}-1}\sim \frac{\alpha +1}{(\alpha q+1)^\frac{1}{q}} \frac{r^\frac{1}{p}r^{\alpha +\frac{1}{q}}}{r^{\alpha+1}}=\frac{\alpha +1}{(\alpha q+1)^\frac{1}{q}} .$$