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Instead of using both PrimeOmega and PrimeNu I'd rather use only SquareFreeQ.

Let's compare appropriate timings:

First @ AbsoluteTiming[ a = PrimeOmega[Range[300000]];
                b = PrimeNu[Range[300000]]; 
                Inner[If[#1 === #2, True, False] &, a, b, List];] 

First @ AbsoluteTiming[SquareFreeQ /@ Range[300000];]

19.748000

 1.521000 

and of course:

Inner[ If[ #1 === #2, True, False] &, a, b, List] == (SquareFreeQ /@ Range[300000])

True

Edit

If we are to find numbers which satisfy PrimeOmega[x] == PrimeNu[x]] in a given range we can use Select, e.g.

Select[ Range[10^6, 10^6 + 11], SquareFreeQ]

{1000001, 1000002, 1000003, 1000005, 1000006, 1000007, 1000009, 

 1000010, 1000011, 1000013, 1000014, 1000015, 1000018, 1000019}

to count them we use:

Count[ Range[10^6, 10^6 + 20], _?SquareFreeQ]

14

of course we might use Count[Range[10^6, 10^6 + 20], _?(PrimeOmega[#] == PrimeNu[#] &)] instead but the latter is slower.
Let's define appropriate counting function:

cf = {#, Count[Range[#, # + 500], _?SquareFreeQ]} & /@ Range[0, 100000, 500];

Namely we count square free numbers in every range: {0, 500}, {500, 1000},...,{99500, 100000}.

Let's plot the counting function cf:

With[{ mcf = Mean @ cf[[All, 2]]}, 
     ListPlot[ cf, AxesOrigin -> {0, 280}, PlotRange -> {280, 320}, AspectRatio -> 1/5, 
               PlotMarkers -> Automatic, Filling -> mcf, 
               Epilog -> {Darker @ Green, Line[{{0, mcf}, {100000, mcf}}]}]]