This is pretty obvious if $H$ contains a Sylow $p$-subgroup of $G$, since then the Sylow $p$-subgroups of $H$ are just the Sylow $p$-subgroups of $G$ contained in $H$, so $n_p(H) \leq n_p(G)$ since $\operatorname{Syl}_p(H) \subseteq \operatorname{Syl}_p(G)$.
If $H$ is smaller though, then something strange seems possible: a Sylow $p$-subgroup of $G$ can have a ton of smaller $p$-subgroups, perhaps $H$ could conspire to have all those smaller $p$-subgroups as Sylow subgroups. For instance, the alternating group of order 12 has only one Sylow 2-subgroup, but it has 3 subgroups of order 2. Why can't a subgroup have those 3 smaller subgroups as its Sylows?
The answer is Lagrange's theorem. If $Q_1, Q_2 \leq P$ are two $p$-subgroups of a Sylow $p$-subgroup $P$ and $Q_1, Q_2 \leq H$ are both contained in $H$, then of course the subgroup $Q=\langle Q_1 ,Q_2 \rangle$ generated by these two $p$-subgroups is contained in both $P$ and $H$. Since $Q \leq P$, it is a $p$-subgroup, and since $Q \leq H$, it is a $p$-subgroup contained in $H$. If $Q_1 \neq Q_2$, then $|Q| > \max(|Q_1|,|Q_2|)$, and so neither $Q_1$ nor $Q_2$ can be a Sylow $p$-subgroup of $H$. In particular, a Sylow $p$-subgroup $P$ of $G$ can only contribute at most one Sylow $p$-subgroup of $H$, namely $H \cap P$.
Not all $H\cap P$ are actually Sylow $p$-subgroups of $H$, but every Sylow $p$-subgroup of $H$ is contained in one of these by Sylow's containment/extension theorem: every $p$-subgroup of $G$ (such as a Sylow $p$-subgroup of $H$) is contained in a Sylow $p$-subgroup of $G$.
I will expand my comment into an answer.
Let $S$ be the set of elements that do not lie in any Sylow $p$-subgroup of $G$. You have shown by a counting argument that $|S|=p$. Let $q$ be any prime that divides $p+1$.Then $S$ must contain some element $g$ of order $q$.
Since $n_p=p+1$, we have $N_G(P) = P$, so $g \not\in N_G(P)$. Let $x$ be a generator of $P$. Then the powers $x,x^2,\ldots, x^{p-1}$ of $x$ all generate $P$, so none of them can centralize $g$. Hence the $p$ elements $\{ g, g^x,g^{x^2}, \ldots, g^{x^{p-1}} \}$ (where $g^h$ means $hgh^{-1}$) are all distinct. Since they all have order $q$, they all lie in $S$, and so $S = \{ g, g^x,g^{x^2}, \ldots, g^{x^{p-1}} \}$.
So every element of $S$ has order $q$, and hence $q$ must be the only prime dividing $p+1$, so $p+1$ is a power of $q$, and $S \cup \{ 1 \}$ must be the unique Sylow $q$-subgroup of $G$.
Best Answer
I like this question, and wanted it to have a bit of a longer answer:
Surprising
This result should be a little surprising. After all the Sylow $p$-subgroups of $H$ can have smaller order, and even though $G$ may have only a few subgroups of order $p^n$, it might have a ton of order $p^{n-1}$. For example, in $G=A_4$, we have only $n_2(G)=1$ Sylow $2$-subgroup, but it has $3$ subgroups of order $2^1$, and so a subgroup $H$ has $n_2(H) \leq 3$, but that leaves open the possibility of $n_2(H) \in \{2,3\}$ both of which contradict the theorem. There is an analogue of $A_4$ called $G=\operatorname{AGL}(1,p^2)$ with approximately the same behavior: $n_p(G)=1$ but $G$ has $p+1$ subgroups of order $p$. When one actually looks for a subgroup $H$, one runs into a problem: no single $H$ contains all those subgroups of order $p$ (or even two of them in the $A_4$ and AGL cases) unless it contains an entire Sylow $p$-subgroup (making those smaller $p$-subgroups irrelevant).
Proof
This idea leads to a simple proof (given by Derek Holt in the comments).
We construction a 1-1 function $f$ from $\operatorname{Syl}_p(H)$ to $\operatorname{Syl}_p(G)$ where $\operatorname{Syl}_p(X)$ is the set of Sylow $p$-subgroups of the finite group $X$. Given a Sylow $p$-subgroup $Q$ of $H$, Sylow's containment theorem says that $Q$ (a $p$-subgroup of $G$) is contained in some Sylow $p$-subgroup $f(Q)$ of $G$. If $f(Q_1) = f(Q_2)$, then $Q_1, Q_2 \leq f(Q_1)$ and $Q_1, Q_2 \leq H$, so $Q_1, Q_2 \leq f(Q_1) \cap H$. However, $f(Q_1) \cap H$ is a $p$-subgroup of $H$, and a $p$-subgroup of $H$ can only contain at most a single Sylow $p$-subgroup of $H$, so $Q_1 = Q_2$. Hence $f$ is a 1-1 function, and $n_p(H) \leq n_p(G)$.
Normal subgroups
If $H$ is a normal subgroup of $G$, then in fact $n_p(H)$ divides $n_p(G)$. Hall (1967) calculated $n_p(G) = n_p(H) \cdot n_p(G/H) \cdot n_p(T)$ where $T=N_{PH}(P \cap H)$ for any Sylow $p$-subgroup $P$ of $G$.
(See also this answer of Mikko Korhonen.)