In general, the answer to this has to be "no". To construct a counterexample, let $S$ be a dense subset of the unit interval (0,1], and let $\mathcal{H}=\ell^2(S)$ be the Hilbert space with orthonormal basis $(e_s)_{s\in S}$. For each positive integer $n$, let $\mathcal{H}_n$ be the subspace of $\mathcal{H}$ generated by linear combinations of $e_s$ over $1/(n+1) < s\le 1/n$. We can define a $C^*$ sub-algebra $\mathcal{A}$ of the bounded linear operators $B(\mathcal{H})$ as follows: $A\in B(\mathcal{H})$ is in $\mathcal{A}$ if and only if the following are satisfied

 1. $A$ maps $\mathcal{H}_n$ into $\mathcal{H}_n$.
 2. There exists a sequence of complex numbers $a_n$ such that, for each fixed $n$,
$$\Vert Ae_s-a_ne_s\Vert+\Vert A^*e_n-\bar a_ne_s\Vert\to0$$
as $s\to 1/n$.

It is easily seen that this is a sub C*-algebra. If operator $A$ is in its center then it must project into the center of each $B(\mathcal{H}_n)$, so must be a constant on each of these. That is, $Ae_s=\lambda_se_s$ for some $\lambda_s\in\mathbb{C}$, which is independent of $S$ over $1/(n+1) < s\le 1/n$. Also, $s\mapsto\lambda_s$ must be continuous at $s=1/n$, from which we see that $s\mapsto\lambda_s$ is constant. So, $\mathcal{A}$ has trivial center.

Fixing a positive integer $m$, define an operator $A\in\mathcal{A}$ by $Ae_s=\min(ms,1)e_s$. Then, $\Vert A\Vert = 1$ and ${\rm dist}(A,{\rm Center}(\mathcal{A}))=1/2$. On the other hand, the restriction of $A$ to each $\mathcal{H}_n$ is within a distance of $1/(2(m+1))$ the center. This gives $\Vert AB - BA\Vert \le \Vert B\Vert / (m+1)$ for all $B\in\mathcal{A}$.

Choosing $m$ as large as we like, this contradicts your claim.