$\because \left(x-\sqrt{x^{2}-2008}\right)\left(y-\sqrt{y^{2}-2008}\right)=2008$，

$\therefore x-\sqrt{x^{2}-2008}=\dfrac{2008}{y-\sqrt{y^{2}-2008}}=y+\sqrt{y^{2}-2008}$，

$y-\sqrt{y^{2}-2008}=\dfrac{2008}{x-\sqrt{x^{2}-2008}}=x+\sqrt{x^{2}-2008}$，

由以上两式可得$x=y$.

$\therefore \left(x-\sqrt{x^{2}-2008}\right)^{2}=2008$，解得：$x^{2}=2008$，

$\therefore 3x^{2}-2y^{2}+3x-3y-2007=3x^{2}-2x^{2}+3x-3x-2007=x^{2}-2007=1$.

故选$D$.