Ligand substitution reaction in coordination compounds
The reaction you wrote down is wrong on two counts. The reactant is not a hypothetical tetraaqua complex and the product is not a hypothetical tetraammin complex. The correct reaction is as shown below:

$$\ce{[Cu(H2O)6]^2+ (aq) + 4 NH3 (aq) &lt;=&gt; [Cu(NH3)4(H2O)2]^2+ (aq) + 4 H2O (l)}\tag{1}$$

Note that I have used an equilibrium arrow here: the reaction can proceed in both directions depending on the concentration of ammonia in your sample (in reality, there is also the competing precipitation of $\ce{Cu(OH)2}$ at slightly basic pH levels but I will ignore that here).

As an equilibrium, this can be pushed either way. So let’s take a look at what nitric acid can do. It is both an acid and an oxidising agent – but as you noted, copper(II) is already very highly oxidised and it is generally not possible to sustain copper(III) or higher copper oxidation states in aquaeous solution (it does exist as an intermediate in various copper-catalysed organic reactions). On the other hand, $\ce{HNO3}$ can very much act as an acid in your solution as there is a basic compound able to receive protons (see spoiler after figuring out the reaction yourself).


   $$\ce{HNO3 (aq) + NH3 (aq) -&gt; NH4+ (aq) + NO3- (aq)}\tag{2}$$


Note that while $(2)$ is also a reversible reaction, the equilibrium is far to the product side so we can essentially treat it as completed. Obviously, this has an effect on the original reaction – and with a simple application of Le Chatelier’s principle the resulting answer should be obvious.


   Yes, since ammonia will be protonated to give ammonium, equilibrium $(1)$ will be shifted back to the reactant side, regenerating hexaaquacopper(II) $\ce{[Cu(H2O)6]^2+}$.