I think you reproduced the "elementary" claim correctly (for the case $n=1$). However, as stated in my earlier comment, the claim is false in general, even if you assume that $p>0$ on $(0,1]$.

Indeed, the claim was that the limit
\begin{equation*}
\lim_{h\downarrow0}r(h) \tag{1}
\end{equation*}
exists and is finite, where
\begin{equation*}
r(h):=\frac{\text{num}(h)}{\text{den}(h)},
\end{equation*}
\begin{equation*}
\text{num}(h):=\int_0^h \ln^{1/2}\frac1u\;dp(u),\quad
\text{den}(h):=p(h)\ln^{1/2}\frac1h.
\end{equation*}
Note that
\begin{equation*}
\text{num}(h)\ge\int_0^h \ln^{1/2}\frac1h\;dp(u)=\text{den}(h),
\end{equation*}
so that $r(h)$ is always $\ge1$.

However, let us show that the limit in (1) can be $\infty$:

Indeed, let
\begin{equation*}
p(u):=\frac1{\sqrt{\ln\frac1u}\,\ln^2\ln\frac1u}
\end{equation*}
for $u\in(0,1/3)$, with $p(0):=0$ and $p(u):=\frac1{\sqrt{\ln3}\,\ln^2\ln3}$ for $u\in[1/3,1]$.
Then $p$ is a real-valued continuous nondecreasing function on $[0, 1]$. Also,
\begin{equation*}
p'(u)\sim\frac1{2u\ln^{3/2}\frac1u\,\ln^2\ln\frac1u}
\end{equation*}
(as $u\downarrow0$) and hence for $h\downarrow0$
\begin{equation*}
\text{num}(h)
\sim\int_0^h \frac{du}{2u\ln\frac1u\,\ln^2\ln\frac1u}
=\frac1{2\ln\ln\frac1h},
\end{equation*}
whereas
\begin{equation*}
\text{den}(h)=\frac1{\ln^2\ln\frac1h},
\end{equation*}
so that
\begin{equation*}
r(h)\sim\frac12\,\ln\ln\frac1h\to\infty.
\end{equation*}

On a somewhat positive note, suppose that $p$ is continuous on $[0,1]$ with $p(0)=0$, has a positive derivative $p'$ in a right neighborhood of $0$, and
\begin{equation}
p(h)=(c+o(1)) hp'(h)\ln\frac1h \tag{2}
\end{equation}
as $h\downarrow0$ for some $c\in[0,2)$. This condition will hold e.g. if (i) $p(h)=Ch^a\ln^b\frac1h$ for some positive real $C$ and $a$, some real $b$, and all small enough $h>0$ or if (ii) $p(h)=C/\ln^t\frac1h$ for some positive real $C$, some real $t>1/2$, and all small enough $h>0$.

Given the additional condition (2), you will have $r(0+)=\frac1{1-c/2}\in(0,\infty)$, by l'Hospital's rule. Indeed, then for the "derivative ratio" we have
\begin{equation*}
\frac{\text{num}'(h)}{\text{den}'(h)}
=\frac{p'(h)\ln^{1/2}\frac1h}{p'(h)\ln^{1/2}\frac1h-\frac{p(h)}{2h}\ln^{-1/2}\frac1h}
=\frac1{1-\dfrac{p(h)}{2hp'(h)\ln\frac1h}}\to\frac1{1-c/2}
\end{equation*}
as $h\downarrow0$.