Removing nodes may or may not improve the quality of the ranking.
- Consider a graph with vertices a, a', b_1\cdots b_{1000}, b'_1\cdots b'_{50}, c_1\cdots c_{1000}. With edges a' \to a, $b_i\to a$. b'_i \to a' and $c_i \to b_i$. If you prune the graph to remove all vertices with no incoming links (that would be $c_i$ and $b'_i$), then the weight on a' would drop compare to $a$. And you may say that this gives "better results".
- On the flip side, consider your initial graph as the a, a', b_1\cdots b_{1000}, b'_1\cdots b'_{50}, c', d' with a' \to a, $b_i \to a$, b'_i \to a', c' \to a' and d' \to c'. PageRank should clearly show that $a$ is the best. But if you remove all vertices with no incoming links, you are left with a,a',c',d' and the ranking of $a$ and a' would get closer.
Basically, the problem is that PageRank is computed recursively. So depending on the exact nature of your graph, how the effect of removing some low scoring nodes propagate can differ. To give another contrived example:
Let your graph be given by $a_1,\ldots, a_{1000}, b_1, \ldots b_{1000}, c_1, \ldots, c_5, d_1, \ldots d_{100}$. Let the arrows go $a_{i+1} \to a_i$ and $b_{i+1}\to b_i$. And let $c_k \to a_{1000}$ and $d_k\to b_{1000}$. Then in this scheme $b_1$ is a better ranked object than $a_1$. But if you remove just 5% of the vertices by removing all of $c_i$ and $d_i$, then PageRank can't tell $a_1$ and $b_1$ apart.