Are you sure, that the character you want to find is a space and not another whitespace character?
If it IS a space, then this would work same as in your example:
fld9.Value = StrRight(Cstr(doc.request(0)), " ")
Or to be more flexible
Dim strSpace as String
strSpace = Chr$(20)
fld9.Value = StrRight(Cstr(doc.request(0)), strSpace)
To find out, what you got there, you take a string (that you probably know) and do this:
Dim strSpaceChar As String
Dim lngAsc as Long
strSpaceChar = Mid$("Your String with a space", 5, 1)
' 5 = Position of the "space" character, 1 = Number characters
lngAsc = Asc( strSpaceChar )
If lngAsc <> 20 then you need to replace strSpace in the example above with the right Character...
If there might be different "spaces" in the text and you do not know before the run, then make an array with all known variations of "space" (the example below would consider NewLines and an HTML nbsp as Spaces)
Dim arrSpaces( 3 ) as String
arrSpaces( 0 ) = Chr$( 10 )
arrSpaces( 1 ) = Chr$( 13 )
arrSpaces( 2 ) = Chr$( 0 )
arrSpaces( 3 ) = Chr$( 160 )
and replace it with a "real" space...
myString = Replace( Cstr(doc.request(0)) , arrSpaces , Chr$(20) )
EDIT because of new kind of question:
There are classes to handle Date / Time values.
You could do something like:
Dim dtRequest as NotesDateTime
Dim strDate as String
Dim strTime as String
Set dtRequest = New NotesDateTime( doc.request(0) )
strDate = dtRequest.DateOnly
strTime = dtRequest.TimeOnly