Answer: 561

Small Hint:

First use $x=20$ to find $f(20)$

Big Hint:

Eliminate alternating terms by deriving a recurrence from $f(x)$ to $f(x-2)$

Solution:

First $f(20)=20^2-f(19)=306.$ Subtracting through two steps gives $$\begin{aligned}f(x)&=x^2-(x-1)^2+f(x-2)\\&=2x-1+f(x-2).\end{aligned}$$ Therefore $$\begin{aligned}f(94)&=306+\sum_{j=11}^{47}(4j-1)\\&=306+4(1073)-37\\&=4561.\end{aligned}$$ The requested remainder is $561.$