Schweizer's reagent from/with CuO?
Since, from what I know, $\ce{CuO}$ doesn't react nor dissolve in aq ammonia, does this experiment suggest a trace amounts of $\ce{Cu(OH)2}$ present in the (assumed) $\ce{CuO}$ precipitate, later forming a low-quality Schweizer's reagent, or am I missing something here?


Well in general this is true, but because the fresh precipitate can be partially hydrated and amorphous, which does allow the ammonium hydroxide to dissolve it.


  
  was the precipitate really $\ce{CuO}$ (if so, how to avoid creating it, i.e. what are the necessary conditions for $\ce{Cu(OH)2}$ creation here), and what is the easiest way to check for $\ce{CuO}$ presence in general?
  


Your precipitate was likely a hydrated copper oxide. Heat causes the copper hydroxide to lose its water.  To prevent $\ce{CuO}$ from forming, you want to keep the solution cool, perhaps with an ice bath. The easiest way to verify this is by visual observation. Copper oxide is black and copper hydroxide is blue, there is no false positive/negative to this for your preparation.


  
  how efficient should be a properly done homemade Schweizer's reagent (using the aforementioned $\ce{CuSO4 + NaOH + 25\%  NH4OH}$ route) in dissolving various celluloses in practice?
  


This is not easy to answer and depends greatly on the concentration of your reagents, temperature, pH and so on. I've seen videos on YouTube however showing about 2g of cotton balls dissolving in 500mL solution as a reference