This appears to be the case, but I was forced to rely on a somewhat complicated inequality on two real variables that looks quite plausible numerically, though I do not have a 100% rigorous proof of it:

**Claim 1**.  Suppose $0 \leq p \leq q \leq 1$ with $p+q \leq 1$ and $h(q) \leq h(p)$, where $h(x) := x \log \frac{1}{x}$.  Let $a = 1/4$ if $q \leq 1/2$, or $a = q(1-q)$ if $q \geq 1/2$.  Then at least one of
$$ a (\log \frac{1}{p} - \log \frac{1}{q}) \leq (q-p) (h(p)+h(q)+h(1-p-q))\tag{$*$}\label{477014_star}$$
and
$$ (1-q) \log \frac{1}{p} \leq \frac{q}{1-q} h(q).\tag{$**$}\label{starstar}$$
holds.

In the triangle where $0 \leq p \leq q \leq 1$ and $p+q \leq 1$, the following plot shows the region where $h(q) \leq h(p)$ (in green) and where \eqref{477014_star} fails (in blue), and where \eqref{starstar} holds (grey), so numerically the intersection of the green and blue regions lie in the grey region.

[![Regions in the claim][1]][1]

Now to explain why the claim gives the result.  We would like to show, for a given $1 \leq k < n$, that the sum of the $k$ largest values of $\frac{h(p_i)}{H(\mathbf{p})}$ is bounded by the sum of the $k$ largest values of $p_i$.  

Let $E$ denote the indices $i$ corresponding to the $k$ largest values of $h(p_i)$.  We also write $\theta := \sum_{i \in E} p_i$, so that $1-\theta = \sum_{j \not \in E} p_j$.

The easy case is when the set $E$ is also the $k$ largest values of $p_i$, thus if $i$ lies in $E$ and $j$ lies outside of $E$ then $\log \frac{1}{p_j} \geq \log \frac{1}{p_i}$.  Averaging, we conclude that
$$ \frac{1}{1-\theta} \sum_{j \not \in E} p_j \log \frac{1}{p_j} \geq \frac{1}{\theta} \sum_{i \in E} p_i \log \frac{1}{p_i}$$
which can be rearranged to
$$ \sum_{i \in E} \frac{h(p_i)}{H(\mathbf{p})} \leq \theta = \sum_{i \in E} p_i,$$
as required.  Note that this includes the case you already treated in which all the $p_i$ are at most $1/e$.

Now we deal with the hard case when $E$ is not the $k$ largest indices of $p_i$.  Let $p$ denote the smallest value of $p_i$ for $i \in E$, and let $q$ be the maximal value of $p_i$, which is attained outside of $E$.  We will try here to estimate all relevant quantities in terms of $p$ and $q$.  

Clearly $1 \leq p \leq q \leq 1$ and $p+q \leq 1$, and also $h(p) \geq h(q)$.  If $i \in E$ and $j \not \in E$, then by construction we have $p_j \leq p_i$ or $p \leq p_i \leq p_j \leq q$, hence in either case
$$ \log \frac{1}{p_j} \geq \log \frac{1}{p_i} - (\log \frac{1}{p} - \log \frac{1}{q}).$$

Averaging, we conclude
$$ \sum_{j \not \in E} \frac{p_j}{1-\theta} \log \frac{1}{p_j} \geq \sum_{i \in E} \frac{p_i}{\theta} \log \frac{1}{p_i} - (\log \frac{1}{p} - \log \frac{1}{q})$$
which one can rearrange as
$$ \sum_{i \in E} p_i \log \frac{1}{p_i} \leq \theta H(\mathbf{p}) + \theta(1-\theta) (\log \frac{1}{p} - \log \frac{1}{q}).$$
Meanwhile, the sum of the $k$ largest values of $p_i$ at least $\theta + (q-p)$, so it suffices to show that
$$ \theta H(\mathbf{p}) + \theta(1-\theta) (\log \frac{1}{p} - \log \frac{1}{q}) \leq (\theta + (q-p)) H(\mathbf{p})$$
or equivalently
$$ \theta(1-\theta) (\log \frac{1}{p} - \log \frac{1}{q}) \leq (q-p) H(\mathbf{p}).$$
Since the $p_i$ contain both $p$ and $q$ as attained by distinct indices, we have
$$ H(\mathbf{p}) \geq p \log \frac{1}{p} + q \log \frac{1}{q} + (1-p-q) \log \frac{1}{1-p-q}.$$
Also, $\theta(1-\theta)$ is bounded by $1/4$, and for $q \geq 1/2$ it is additionally bounded by $q(1-q)$ since $\theta \leq 1-q$.  The result then follows if \eqref{477014_star} holds.

To handle the remaining case we make an alternate estimate of the various quantities involved in terms of $p$ and $q$.  Clearly
$$ H({\mathbf p}) \geq \sum_{i \in E} h(p_i) + h(q)$$
so as the sum of the $k$ largest probabilities is certainly at least $q$, it would suffice to show that
$$ \sum_{i \in E} h(p_i) \leq q (\sum_{i \in E} h(p_i) + h(q))$$
or equivalently
$$ \sum_{i \in E} h(p_i) \leq \frac{q}{1-q} h(q).$$
Since $h(p_i) \leq p_i \log \frac{1}{p}$ for all $i \in E$, and $\sum_{i \in E} p_i \leq 1-q$, it thus suffices to have
$$ (1-q) \log \frac{1}{p} \leq \frac{q}{1-q} h(q).$$
which is \eqref{starstar}.  So the result follows from Claim 1.
 


  [1]: https://i.sstatic.net/XIOFHNlc.png