No, it is only a coideal subalgebra. The subalgebra bit is clear (I hope).
Let us see why it is a coideal. This is a general coalgebraic fact. I keep your letters but assume they are all coalgebras only. I pick a basis of $F$ consisting of three types of elements:
- $a_i$ - a basis of the kernel $\ker(\varphi)$,
- $b_j$ - extend it to a basis of $\varphi^{-1}(H)$,
- $c_k$ - extend it to a basis of $F$. Now pick $x\in \varphi^{-1}(H)$. What do we know about $\Delta (x)$? We know that $\varphi\otimes\varphi (\Delta (x))= \Delta (\varphi (x)) \in H \otimes H$. This means $\Delta (x)$ is a linear combination of the following basis elements of $F\otimes F$: $$a_i\otimes c_j, \ a_i \otimes b_j, \ a_i \otimes c_j,\ b_i \otimes a_j,\ b_i \otimes b_j,\ c_i \otimes a_j.$$ All of them belong to $F\otimes \varphi^{-1}(H) + \varphi^{-1}(H)\otimes F$, so it is a coideal.
It remains to see why it is not a subcoalgebra. The example by Ostrik in the comments is perfect. Let $F=kF_1$ and $G=kG_1$ be group algebras. All its Hopf subalgebras are group subalgebras and the homomorphisms are $\varphi =k \phi$ for a group map $\phi$. Now pick a subgroup $H_1 \leq G_1$, then $$\varphi^{-1} (H)= k\phi^{-1}(H_1) + \mbox{Span} \{ x-xy \,\mid\, x \in F_1, y \in \ker(\phi)\}.$$ This is a Hopf subalgebra if and only if $\phi$ is injective or $H_1=G_1$.