I got $$r([A,B])\leq 4\sqrt{2} r(A) r(B).$$
It is lower than $8$ but still higher than the conjuncture $C_{nr}=4$.
I used the following facts:
For normal (i.e. $X^*X=XX^*$) matrices we have $r(X)=\sigma_1(X)$ (the largest singular value of X).
$\sigma_1(XY-YX)\leq 2 \sigma_1(X)\sigma_1(Y)$
Also, note that for $X$ and $Y$ hermitian (i.e. $X^*= X$ and $Y^* = Y$ ) we have
$$r(X+iY) \geq \max\left(\sigma_1(X),\sigma_1(Y)\right),$$ $$r(X+iY) \leq \sqrt{\sigma_1^2(X) + \sigma_1^2(Y)}.$$
Lets decompose $A$ and $B$ in their hermitian and antihermitian parts, $$A = A_h + i A_a, \quad B= B_b+i B_h+i B_a.$$
Then $$r^2([A,B]) \leq \left( \sigma_1([A_h,B_h]-[A_a,B_a])\right)^2 + \left( \sigma_1([A_h,B_a]-[A_a,B_h])\right)^2$$ sigma_1([A_h,B_a]+[A_a,B_h])\right)^2$$ $$\leq\left( 2\sigma_1(A_h)\sigma_1(B_h)+2\sigma_1(A_a)\sigma_1(B_a) \right)^2 + \left( 2\sigma_1(A_h)\sigma_1(B_a)+2\sigma_1(A_a)\sigma_1(B_h) \right)^2$$ $$=4 ( \sigma_1^2(A_h) + \sigma_1^2(A_a) )( \sigma_1^2(B_h) + \sigma_1^2(B_a) ) +16 \sigma_1(A_h) \sigma_1(A_a) \sigma_1(B_h) \sigma_1(B_a)$$ $$\leq 8( \sigma_1^2(A_h) + \sigma_1^2(A_a) )( \sigma_1^2(B_h) + \sigma_1^2(B_a) )$$ $$\leq 32 r^2(A)r^2(B).$$
