EDIT: I didn't consider multiple dates in my approach. See Yarnabrina's post below for a more robust solution.
You rightly identified tidyr::complete() as the most appropriate function for the job. It works for all types of values not just dates. All you need to do is supply it with the sequence (increments of 100 in this case).
library(tidyverse)
df <- tribble(~timecode, ~value,
201901010000, 3.5,
201901010100, 4.2,
201901010200, 3.2,
201901010300, 1.5,
201901010400, 5.3,
201901010800, 3.8,
201901010900, 2.9,
201901011000, 4.6)
complete(df, timecode = seq(min(timecode), max(timecode), by = 100))
#> # A tibble: 11 x 2
#> timecode value
#> <dbl> <dbl>
#> 1 201901010000 3.5
#> 2 201901010100 4.2
#> 3 201901010200 3.2
#> 4 201901010300 1.5
#> 5 201901010400 5.3
#> 6 201901010500 NA
#> 7 201901010600 NA
#> 8 201901010700 NA
#> 9 201901010800 3.8
#> 10 201901010900 2.9
#> 11 201901011000 4.6
Created on 2020-07-03 by the reprex package (v0.3.0)