The answer is no. Indeed, if this conjecture were true, then the conditions $f(a)=g(a)$, $f(b)=g(b)$, and 
\begin{equation}
	k_f(x):=\frac{|f''(x)|}{(1+f'(x)^2)^{3/2}}=k_g(x) 
\end{equation}
for all $x \in [a, b]$ would imply that 
\begin{equation}
	\ell_f:=\int_a^b\sqrt{1+f'(x)^2}\,dx=\ell_g; 
\end{equation}
that is, we would have that the length of the graph of a smooth enough function $f$ over $[a,b]$ would be determined by the curvature of the graph and the values of $f$ at $a,b$. 

Let us show that this is not so. Indeed, let 
\begin{equation}
	h(x):=\frac12 \sqrt{\frac{(1+4 x^2)^{3/2}-3 x-8 x^3}{x+3 x^3}}. 
\end{equation}
Everywhere here $x>0$, unless otherwise specified. 
It is not hard to see that the expression under the square root is $>0$, so that the definition of $h$ is correct. 
Let now 
\begin{equation}
	f(x):=\int_a^x h(u)\,du
\end{equation}
and 
\begin{equation}
	g(x):=x^2-a^2,
\end{equation}
where $a:=1/100$. 
Then 
\begin{equation}
	\frac{f''(x)}{(1+f'(x)^2)^{3/2}}=\frac{-2}{(1+4x^2)^{3/2}}=-\frac{g''(x)}{(1+g'(x)^2)^{3/2}}. 
\end{equation}
So, the function $f$ is concave, the function $g$ is convex, $f(a)=g(a)$, and 
\begin{equation}
	k_f=k_g. 
\end{equation}
Moreover, $f'(a)>g'(a)$. So, there is a unique real $b>a$ such that $f(b)=g(b)$; in fact, $b=0.710526\ldots\;$. 
However, 
\begin{equation}
	\ell_f=0.920463\ldots\ne0.894858\ldots=\ell_g,
\end{equation}
which does disprove the conjecture. 

This counterexample is illustrated here:

[![enter image description here][1]][1]


  [1]: https://i.sstatic.net/ybM1Z.png