Solution:
(Ⅰ) According to the problem, the distance from the moving point $P(x,y)$ ($x \geqslant 0$) to the fixed point $F(1,0)$ is equal to the distance from point $P(x,y)$ to the line $x=-1$,
By the definition of a parabola, the trajectory equation of point $P$ is a parabola with focus $F(1,0)$ and directrix $x=-1$,
Therefore, the equation of curve $C$ is $y^{2}=4x$.
(Ⅱ) Since $D(x_{0},2)$ is on curve $C$, we get $4=4x_{0} \Rightarrow x_{0}=1$, thus $D(1,2)$.
Let $M(x_{1},y_{1})$, $N(x_{2},y_{2})$,
Line $l_{1}$: $y=k(x-1)+2$,
Then $l_{2}$: $y=-k(x-1)+2$,
From $\begin{cases} y=k(x-1)+2 \\ y^{2}=4x \end{cases} \Rightarrow k^{2}x^{2}-(2k^{2}-4k+4)x+(k-2)^{2}=0$,
$\therefore x_{1} \times 1= \dfrac {(k-2)^{2}}{k^{2}}= \dfrac {k^{2}-4k+4}{k^{2}}$
Similarly, $x_{2}= \dfrac {k^{2}+4k+4}{k^{2}}$,
$\therefore x_{1}+x_{2}= \dfrac {2k^{2}+8}{k^{2}}, x_{1}-x_{2}= \dfrac {-8}{k}$,
$\therefore y_{1}-y_{2}=k(x_{1}+x_{2})-2k= \dfrac {8}{k}$
$\therefore k_{MN}= \dfrac {y_{1}-y_{2}}{x_{1}-x_{2}}= \dfrac { \dfrac {8}{k}}{- \dfrac {8}{k}}=-1$
The slope of line $MN$ is a constant $-1$. Therefore, the final answers are:
(Ⅰ) The equation of curve $C$ is $\boxed{y^{2}=4x}$.
(Ⅱ) The slope of line $MN$ is a constant $\boxed{-1}$.