This question is rather old, but is a good question and deserves an answer.
Yes! textrank returns all the information that you need. Just look
at the output of str(tr). Part of it says:
$ sentences_dist:Classes ‘data.table’ and 'data.frame': 666 obs. of 3 variables:
..$ textrank_id_1: int [1:666] 1 1 1 1 1 1 1 1 1 1 ...
..$ textrank_id_2: int [1:666] 2 3 4 5 6 7 8 9 10 11 ...
..$ weight : num [1:666] 0.1429 0.4167 0 0.0625 0 ...
This gives which sentences are connected in the form of a lower triangular matrix. Two sentences are connected if the weight of their connection is greater than zero. To visualize the graph, use the non-zero weights as an edgelist and build the graph.
Links = which(tr$sentences_dist$weight > 0)
EdgeList = cbind(tr$sentences_dist$textrank_id_1[Links],
tr$sentences_dist$textrank_id_2[Links])
library(igraph)
SGraph1 = graph_from_edgelist(EdgeList, directed=FALSE)
set.seed(42)
plot(SGraph1)

We see that 11 of the nodes (sentences) are not connected to any other node.
For example, sentences 15 and 36
tr$sentences$sentence[c(36,15)]
[1] "Contact:"
[2] "Integration of the models into the existing architecture."
But other other nodes do connect up, for example node 1 is connected to node 2.
tr$sentences$sentence[c(1,2)]
[1] "Statistical expert / data scientist / analytical developer"
[2] "BNOSAC (Belgium Network of Open Source Analytical Consultants),
is a Belgium consultancy company specialized in data analysis and
statistical consultancy using open source tools."
because those sentences share the (important) words "statistical", "data", and "analytical".
The singleton nodes take up a lot of space in the graph making the other nodes rather crowded. So I will also show the graph with those removed.
which(degree(SGraph1) == 0)
[1] 4 7 15 20 21 23 25 26 29 30 36
SGraph2 = delete.vertices(SGraph1, which(degree(SGraph1) == 0))
set.seed(42)
plot(SGraph2)

That shows the relations between sentences somewhat better, but I expect that you can find a nicer layout for the graph that better shows the relations. However, that is not the thrust of the question and I leave it to you to make the graph pretty.