Is finding a path with more red vertices than blue vertices NP-hard?
Your solution does not work because Dijkstra and Bellman-Ford cannot interpret "simple path" feature. And they will indeed fall in any negative cycle.

I think the best to show NP-completeness, is to use the Hamiltonian path problem. Let's take a graph $G'$ of $N$ red vertices.

Then you build a graph $G$, adding $s$, $t$ and $N-1$ blue vertices to $G'$. You first chain with edges all the blues vertices from the source to the last blue one ($s$->$b_1$->$b_2$->...->$b_{N-1}$). Then you put edges from $b_{N-1}$ to every red vertex and an edge from every red vertex to $t$.

So a single path from $s$ to $t$ passes necessarly through all blue nodes ($N-1$) and then have to pass to all red nodes ($N$) to answer to

Is there a simple path in $G$ from $s$ to $t$ with more red than blue vertices ?

which is thus like answer to:

Is there an Hamiltonian path in $G'$

So your problem is indeed NP-complete.