The answer is no, assuming that you seek an orientation preserving square root.

(I see unknown's answer appeared while I'm typing. I don't quite understand it at the moment but the construction looks different.)

Consider a diffeomorphism $f$ such that $f(0)=0$, $f(1)=1$, $f(t)>t$ for all $t\in(0,1)$, $f'(0)=a>1$, $f'(1)=b<1$ and furthermore $f$ is linear in a neighborhood of 0 and in a neighborhood of 1. Let $g$ be a $C^1$ square root of $f$. Then obviously $g(0)=0$, $g'(0)=\sqrt a$ and moreover $g(t)=\sqrt a\cdot t$ in the neighborhood of 0 where $f^{-1}$ is linear. Indeed, suppose that $g(t_0)=ct_0$ where $c\ne\sqrt a$, then $$ g(a^{-k}t_0) = g(f^{-k}(t_0)) = f^{-k}(g(t_0)) = ca^{-k}t_0 . $$ Since $g(0)=0$ and $g(a^{-k}t_0)=ca^{-k}t_0$, there exist a point $t$ between $0$ and $a^{-k}(t_0)$ (and hence arbitrarily close to 0) such that $g'(t)=c$, contrary to continuity of $g'$ at 0.

Similarly, $g(1)=1$, $g'(1)=\sqrt b$ and $g$ is linear near 1: $g(t)=1+\sqrt b(t-1)$. So we know $g$ near the endpoints. Furthermore there is a compatibility condition: take $t_0$ close to 0 and a very large $n$ such that $f^n(t_0)$ is close to 1. Then $$ f^n(\sqrt a t_0) = f^n(g(t_0)) = g(f^n(t_0)) = 1+ \sqrt b(f^n(t_0)-1) $$ Obviously a generic $f$ does not satisfy this (and hence doest not have a square root), since the $f$-orbits of $t_0$ and $\sqrt a t_0$ are essentially independent.

More formally, modify $f$ near some point $t'=g^{2k+1}(t_0)$ which is far from 0 and 1, so that the resulting function $\tilde f$ equals $f$ outside the segment $[f^{k}(t_0),f^{k+1}(t_0)] = [g^{2k}(t_0),g^{2k+2}(t_0)]$ and $\tilde f(t')\ne f(t')$. The new function satisfies $\tilde f^n(t_0)=f^n(t_0)$ but $$ \tilde f^n(\sqrt a t_0) \ne f^n(\sqrt a t_0) = 1+ \sqrt b(\tilde f^n(t_0)-1) , $$ hence $\tilde f$ cannot have a $C^1$ square root.

A similar argument shows that diffeomorphisms without square root are dense among diffeomorphisms with at least two fixed points.

The answer is no, assuming that you seek an orientation preserving square root.

(I see unknown's answer appeared while I'm typing. I don't quite understand it at the moment but the construction looks different.)

Consider a diffeomorphism $f$ such that $f(0)=0$, $f(1)=1$, $f(t)>t$ for all $t\in(0,1)$, $f'(0)=a>1$, $f'(1)=b<1$ and furthermore $f$ is linear in a neighborhood of 0 and in a neighborhood of 1. Let $g$ be a $C^1$ square root of $f$. Then obviously $g(0)=0$, $g'(0)=\sqrt a$ and moreover $g(t)=\sqrt a\cdot t$ in the neighborhood of 0 where $f^{-1}$ is linear. Indeed, suppose that $g(t_0)=ct_0$ where $c\ne\sqrt a$, then $$ g(a^{-k}t_0) = g(f^{-k}(t_0)) = f^{-k}(g(t_0)) = ca^{-k}t_0 . $$ Since $g(0)=0$ and $g(a^{-k}t_0)=ca^{-k}t_0$, there exist a point $t$ between $0$ and $a^{-k}(t_0)$ (and hence arbitrarily close to 0) such that $g'(t)=c$, contrary to continuity of $g'$ at 0.

Similarly, $g(1)=1$, $g'(1)=\sqrt b$ and $g$ is linear near 1: $g(t)=1+\sqrt b(t-1)$. So we know $g$ near the endpoints. Furthermore there is a compatibility condition: take $t_0$ close to 0 and a very large $n$ such that $f^n(t_0)$ is close to 1. Then $$ f^n(\sqrt a t_0) = f^n(g(t_0)) = g(f^n(t_0)) = 1+ \sqrt b(f^n(t_0)-1) $$ Obviously a generic $f$ does not satisfy this (and hence doest not have a square root), since the $f$-orbits of $t_0$ and $\sqrt a t_0$ are essentially independent.

More formally, modify $f$ near some point $t'=g^{2k+1}(t_0)$ which is far from 0 and 1, so that the resulting function $\tilde f$ equals $f$ outside the segment $[f^{k}(t_0),f^{k+1}(t_0)] = [g^{2k}(t_0),g^{2k+2}(t_0)]$ and $\tilde f(t')\ne f(t')$. The new function satisfies $\tilde f^n(t_0)=f^n(t_0)$ but $$ \tilde f^n(\sqrt a t_0) \ne f^n(\sqrt a t_0) = 1+ \sqrt b(\tilde f^n(t_0)-1) , $$ hence $\tilde f$ cannot have a $C^1$ square root.

A similar argument shows that diffeomorphisms without square root are dense among diffeomorphisms with at least two fixed points.

EDIT: If $f$ has exactly two fixed points, then it does not have an orientation reversing square root as well. Because such a square root $g$ would have exactly one fixed point, and all other fixed points of $g^2$ come in pairs of the form $\{t,g(t)\}$. So $g^2$ must have an odd number of (or infinitely many) fixed points.