The answer is no to both questions. E.g., suppose that $T=2$ and $$f_n(t)=1(t\le1-\tfrac1n)+n(1-t)1(1-\tfrac1n<t\le1).$$ Then $f_n(t)$ increases to $f(t)=1(t<0)$, and all the other conditions on $f_n$ hold.
However, for $t_n:=1-\tfrac1{2n}$ and any strictly increasing continuous function $h\colon[0,2]\to[0,2]$
$$d(f_n,f)\ge|f_n(t_n)-f(h(t_n))|=|\tfrac12-f(h(t_n))|
\ge\tfrac12,$$
since $f(t)\in\{0,1\}$ for each $t\in[0,2]$. So, $d(f_n,f)\not\to0$.