As I mentioned in the comments, I'm hoping to find a one-liner based on Cauchy-Schwartz, as suggested by user Mikael
de la Salle. However, while this is not
available, let me present a proof I just found based on extending $\phi$ to the unitization of $A$.
Let
$$
S = \sup\big \{\|\phi(a)\|: a\geq 0,\ \|a\|\leq 1\big \},
$$
and let
$\Phi $ be the extension of $\phi$ to the unitization $\tilde A$ defined by setting $\Phi (1)=SI_H$.

I claim that $\Phi $ is positive. To see this we must check that
$$
\Phi \big ((a-\lambda 1)^*(a-\lambda 1)\big )\geq 0,
\tag{$\star$}
$$
for every $a$ in $A$, and every $\lambda \in {\mathbb C}$. The case $\lambda =0$ is clearly true, so we may assume that $\lambda \neq 0$. In the latter
case, we may change variables by replacing $a$ with $\lambda a$, and then
divide everithing by $|\lambda |^2$, leading to the following equivalent form of $(\star)$:
$$
\Phi \big ((a-1)^*(a-1)\big )\geq 0,
\tag{$\star\star$}
$$
Observing that
$$
0\leq (a-1)^*(a-1) = a^*a-a^*-a+1,
$$
and fixing an approximate identity $\{u_i\}_i$ for $A$, we have for all $i$ that
$$
u_i(a^*+a -a^*a)u_i \leq u_i^2,
$$
so
$$
\phi\big (u_i(a^*+a -a^*a)u_i\big ) \leq \phi(u_i^2) \leq \|\phi(u_i^2)\|I_H \leq SI_H = \Phi (1).
$$
Taking the limit as $i\to \infty $, the above yields
$$
\phi (a^*+a -a^*a) \leq \Phi (1),
$$
which is equivalent to $(\star\star)$, proving the claim.

By [1, Theorem 1.3.3], (which Størmer proves using Russo-Dye and Cauchy-Schwartz),
we then deduce that
$$
S = \|\Phi (1)\| = \|\Phi \| \geq \|\phi\| \geq S,
$$
concluding the proof.

[1] Størmer, Erling, Positive linear maps of operator algebras, Springer 2013.

3more comments