How to define column that doesn't have NAs in sequence of certain rows?

I'd use the rle() function to count run lengths of !is.na(). For example, using your definition of df1:

df2 <- data.frame(Name = character(3), Group = character(3))

for (i in 1:3) {
  runs <- rle(!is.na(df1[, i + 1]))
  good <- which(runs$values == TRUE)
  runs <- runs$lengths
  n <- length(runs)
  df2$Group[i] <- if (n %in% good) "Stable"
                  else if (max(runs[good]) >= 5) "Was_Stable"
                  else "Not_Stable"
  df2$Name[i] <- names(df1)[i + 1]
}

You can use sapply to iterate over columns and check with any :

sapply(df1[-1], function(x) any(with(rle(!is.na(x)), values & lengths >= 5)))

# Client1 Client2 Client3 
#   TRUE   FALSE    TRUE 

Similar to user2554330, then you can use rle like this:

# add a column with no NAs as an example
df1 <- cbind(df1, dummy = 1:NROW(df1)) 

# find columns with five or more NAs in a row
is_num <- vapply(df1, is.numeric, TRUE) # assume we only look at numerics?
res <- setNames(rep(TRUE, NCOL(df1)), colnames(df1))
res[is_num] <- vapply(df1[is_num], function(x){
  o <- rle(!is.na(x))
  any(o$lengths[o$values] > 4)
}, TRUE)
res
#R> Date Client1 Client2 Client3   dummy 
#R> TRUE    TRUE   FALSE    TRUE    TRUE

I would expect this to be quite fast. If you do not care about the other columns, then you can do:

is_num <- vapply(df1, is.numeric, TRUE)
vapply(df1[is_num], function(x){
  o <- rle(!is.na(x))
  any(o$lengths[o$values] > 4)
}, TRUE)
#R> Client1 Client2 Client3 
#R>    TRUE   FALSE    TRUE

I realise after the fact that this is only a minor twist to the answer by Ronak Shah. Combining his approach with my solution yields this:

vapply(df1[vapply(df1, is.numeric, TRUE)], function(x)
  with(rle(!is.na(x)), any(lengths[values] > 4)), TRUE)
#R> Client1 Client2 Client3 
#R>    TRUE   FALSE    TRUE