The earlier pseduo-code makes two critical assumptions, namely:

1 - a HYPERLINK function exists within each cell specified in standard form - ie not embedded as part of a bigger function

2 - That no comma shall appear within the Hyperlink address itself literally

The below:

Dim rngCell As Range, strLink As String
For Each rngCell In Range("R5:R207")
    With rngCell
        If Left(.Formula, 10) = "=HYPERLINK" Then
            strLink = .Parent.Evaluate(Split(.Formula, ",")(0) & ")")
            Debug.Print strLink
        End If
    End With
Next rngCell
will test for a basic HYPERLINK function but won't handle nuances should the function be contain embedded functions nor will it account for literal commas appearing within the hyperlink address.

If this remains an issue then I would suggest posting a representative sample file such that people can test.

Regards what the code does - the code provided simply generates (or attempts to generate) the requisite URL to a string variable - strLink
What you choose to do with the variable is of course up to you - for the sake of validation I added a Debug.Print line to the above so as to write results to the Immediate Window in VBE