Hard water filtration
To answer your question, what if we did utilize the method you propose?
Before going into the details, it should be noted that the Ion-Exchange method is used to remove hardness of water at an industrial scale. The principal reason that this method is preferred over any other hardness removal process is that it has a unique advantage:
All exchanges take place in an aqueous medium in the form of ions alone.
Keeping that in mind, let's come back to the adjustment you suggested. In the Anion Exchanger, the usual reactions are
$$\ce{R'OH- +Cl^- -&gt; R'Cl- + OH-}$$
$$\ce{2R'OH- + SO4^2- -&gt; R'2 SO4^2- + 2OH-} $$
$$\ce{2R'OH- + CO3^2- -&gt; R'2 CO3^2- + 2OH- }$$
Given that you still haven't removed the cations, the ensuing process would result as follows:
$$\ce{Ca^2+ + OH- &lt;=&gt;&gt; Ca(OH)2} \   \\K_{-1}=\frac{1}{K_{spCa(OH)_2}}$$ and $$\ce{Mg^2+ + OH- &lt;=&gt;&gt;Mg(OH)2}\\K_{-1}=\frac{1}{K_{spMg(OH)_2}}$$
Note that $K_{spMg(OH)_2}=1.8\cdot10^{-11}$ and  ${K_{spCa(OH)_2}}=6\cdot 10^{-6} $
. Further, due to the high concentration of $\ce{OH-}$ ions in the medium, the equilibrium shifts forward, meaning that a high amount of salt is precipitated and deposited within the exchanger, also called scaling. This is detrimental to the industry from any perspective.
Now, let's go back to the part where we discussed advantage of the catex. Do you get it now?
In simpler terms, the fundamental phenomenon behind this process (ION EXCHANGE) is challenged and its efficacy is vastly downgraded.
For future references:

Detailed exchange process
$K_{sp}$ values