I think this will work, but it will take some time to write out a full proof.
(See the update below.)
Let $\phi \colon [0, 1] \to [0, 1]$ be the Cantor function.
For real $a, b, k,$ and $h > 0,$ consider this function ($[b, b + k]$ means $[b + k, b]$ if $k < 0$):
$$
\phi^* = \Phi(a, h, b, k) \colon [a, a + h] \to [b, b + k], \ x \mapsto b + k\phi\left(\frac{x - a}h\right).
$$
I will take for granted, without proof for the moment, that $\phi^*$ is differentiable almost everywhere, with derivative zero where it is defined (this follows from well-known properties of $\phi,$ and so it needn't be proved here) and (I expect this to be not too hard to prove, perhaps by expressing $\phi^*$ as a uniform limit of "step-like" functions based on approximations to the Cantor set) the graph of $\phi^*$ is rectifiable, with arc length $h + |k|.$ (According to the Wikipedia article, this is known to be true when $h = k.$ It is intuitively quite clear why this is so, and the proof ought to generalise to the case of distinct $h, k.$)
For $n = 0, 1, 2, \ldots,$ let $s_n$ be the $n^\text{th}$ partial sum of the series:
$$
1 - \frac12 + \frac13 - \frac14 + \cdots = \log2.
$$
Construct a continuous function $f \colon [0, 1] \to [0, 1]$ by glueing together these functions, for $k = 0, 1, 2, \ldots$:
\begin{align*}
\Phi\left(1 - \frac1{2^{2k}}, \frac1{2^{2k+1}}, s_{2k}, \frac1{2k+1}\right) & \colon
\left[1 - \frac1{2^{2k}}, 1 - \frac1{2^{2k+1}}\right] \to [s_{2k}, s_{2k+1}], \\
\Phi\left(1 - \frac1{2^{2k+1}}, \frac1{2^{2k+2}}, s_{2k+1}, -\frac1{2k+2}\right) & \colon \left[1 - \frac1{2^{2k+1}}, 1 - \frac1{2^{2k+2}}\right] \to [s_{2k+2}, s_{2k+1}],
\end{align*}
where:
\begin{gather*}
f\left(1 - \frac1{2^{n}}\right) = s_n \quad (n = 0, 1, 2, \ldots), \\
f(1) = \log2.
\end{gather*}
Then $f$ is differentiable almost everywhere, with derivative zero wherever it is defined, therefore:
[As Paramanand Singh pointed out in the comments, and as I've only slowly come to understand, the expression on the left cannot be understood as a Riemann integral, therefore my answer does not strictly meet the terms of the question. (See the second update, below.)]
$$
\int_0^1\sqrt{1 + f'(x)^2}\,dx = 1.
$$
But for $n = 1, 2, 3, \ldots,$ the graph of the restriction of $f$ to the interval $[0, 1 - 2^{-n}]$ has arc length:
$$
1 - \frac1{2^n} + \left(1 + \frac12 + \frac13 + \cdots + \frac1n\right)
$$
and this is unbounded, therefore the graph of $f$ is not rectifiable.
Update
It turns out to be remarkably easy to prove that the arc length of the graph of $\phi^*$ is $h + |k|.$
@user856's beautifully simple answer to Arc length of the Cantor function says everything that is really needed, but it can be misunderstood, as can be seen from one of the comments on it. The same reservation applies to Dustan Levenstein's brief comment on Elementary ways to calculate the arc length of the Cantor function (and singular function in general), which I believe is a version of the same argument. In the hope of being easily understood, I will labour the proof. I'm sorry!
For $n = 1, 2, 3, \ldots,$ the $n^\text{th}$ stage of the traditional "middle third" construction of the Cantor set yields $m = 2^n - 1$ pairwise disjoint open intervals, the smallest of which have length $\left(\frac13\right)^n,$ and whose lengths sum to $1 - \left(\frac23\right)^n.$ Given $\epsilon > 0$ with $\epsilon < 2h,$ take $n$ large enough that $\left(\frac23\right)^n < \frac{\epsilon}{2h}.$
Arrange the open intervals in ascending order as $J_1, J_2, \ldots, J_m.$
Set $q_0 = 0, p_m = 1.$ In $J_i,$ for $i = 1, 2, \ldots, m,$ take a closed subinterval $[p_{i-1}, q_i],$ where:
$$
q_i - p_{i-1} \geqslant \left(1 - \frac{\epsilon}{2h}\right)|J_i|.
$$
Construct a polygonal chain $Q_0P_0Q_1P_1\cdots Q_mP_m$ of points on the graph of $\phi^*,$ where:
$$
P_i = (a + hp_i, b + k\phi(p_i)),\quad Q_i = (a + hq_i, b + k\phi(q_i)) \qquad (i = 0, 1, \ldots, m).
$$
Because $\phi$ is constant on each of $J_1, J_2, \ldots, J_m,$ and because in particular $\phi(p_{i-1}) = \phi(q_i)$ for $i = 1, 2, \ldots, m,$ the length of the chain is:
\begin{align*}
& \phantom{={}} \sum_{i=1}^m\|P_{i-1}Q_i\| + \sum_{i=0}^m\|Q_iP_i\| \\
& = \sum_{i=1}^mh(q_i-p_{i-1}) + \sum_{i=0}^m\sqrt{h^2(p_i-q_i)^2 + k^2(\phi(p_i)-\phi(q_i))^2} \\
& > h\sum_{i=1}^m(q_i-p_{i-1}) + |k|\sum_{i=0}^m (\phi(p_i)-\phi(q_i)) \\
& > h\left(1 - \frac{\epsilon}{2h}\right)\sum_{i=1}^m|J_i| + |k|(\phi(p_m) - \phi(q_0)) \\
& > h\left(1 - \frac{\epsilon}{2h}\right)^2 + |k|(\phi(1) - \phi(0)) \\
& > h + |k| - \epsilon.
\end{align*}
I hope it will be clear from the Triangle Inequality - without me labouring the details in the same way - that the length of any chain $Q_0R_1R_2\cdots R_lP_m$ of successive points on the graph of $\phi^*$ is at most $h + |k|.$
It follows that the arc length of the graph of $\phi^*,$ defined as the least upper bound of the lengths of all such chains, is well-defined and is equal to $h + |k|.$
Second update
I'll try to explain in gory detail what has been confusing me so much,
to reduce the risk of confusing others! The function
$g(x) = \sqrt{1 + f'(x)^2}$ is defined, and has the constant value
$1,$ on an open set $E \subset [0, 1],$ whose complement (a union of
countably many scaled, translated copies of the Cantor set) has
measure $0.$ Therefore, any extension of $g$ to the whole of
$[0, 1]$ is Riemann integrable, and the value of any such integral
is $1.$ It does not follow that $g$ itself is Riemann
integrable on $[0, 1]!$ There is simply no definition of the Riemann
integral that applies here.
As far as I can tell, the best that can be done using the Riemann
integral is to apply e.g. section 11.2 of Vladimir A. Zorich,
Mathematical Analysis II (first edition 2004), according to which
$E$ is an "admissible set", and:
$$
\int_E\sqrt{1 + f'(x)^2}\,dx= 1.
$$
This is a proper Riemann integral (Zorich also gives a definition of
an improper Riemann integral, which adds nothing here), but I find
this little consolation.
Best Answer
Short answer.
No there is no such $\gamma$ satisfying those 3 conditions.
Long Answer.
Usually, theorems are true in much more general circumstances than you may have been exposed to, and that is indeed the case here. For example, the following statement is true:
Here, we're working with a proper Riemann-integral on the RHS. Even though the statement involves only Riemann integrals, I do not know how to prove this without more machinery involving absolute continuity/Lebesgue integrals, and especially the Lebesgue version of the FTC. In the above theorem, proving rectifiability of $\gamma$ and the inequality $\leq$ are elementary. The tough part is proving the inequality $\geq$.
Let me just mention how the proof goes. First, define the total variation function $\Gamma:[a,b]\to[0,\infty]$ as \begin{align} \Gamma(x):=L(\gamma|_{[a,x]}). \end{align} Now, let $\{t_0,\dots, t_N\}$ be an arbitrary partition of $[a,x]$; then \begin{align} \sum_{i=1}^N|\gamma(t_{i})-\gamma(t_{i-1})|&=\sum_{i=1}^N\left|\int_{t_{i-1}}^{t_i}\gamma'(t)\,dt\right|\tag{FTC}\\ &\leq \sum_{i=1}^N\int_{t_{i-1}}^{t_i}|\gamma'(t)|\,dt\\ &=\int_a^x|\gamma'(t)|\,dt \end{align} Thus, by taking the supremum over all the partitions, we have that $\Gamma(x):=L(\gamma|_{[a,x]})\leq \int_a^x|\gamma'(t)|\,dt$. In particular, this is true for $x=b$, thus it proves the rectifiability of $\gamma$ and the inequality $\leq$ mentioned in the theorem (in particular, it also shows $\Gamma$ is real-valued; i.e doesn't take the value $\infty$).
To prove the reverse inequality, let us temporarily make the extra hypothesis that $\Gamma$ is differentiable on $[a,b]$ and $\Gamma'$ is Riemann-integrable on $[a,b]$. Then, for any $t\in (a,b)$ and sufficiently small $h>0$, we have by the additivity of lengths of paths that \begin{align} |\gamma(t+h)-\gamma(t)|&\leq L(\gamma|_{[t,t+h]})\\ &=L(\gamma|_{[a,t+h]})-L(\gamma|_{[a,t]})\\ &=\Gamma(t+h)-\Gamma(t) \end{align} Dividing by $h$ and taking the limit as $h\to 0^+$ shows that $|\gamma'(t)|\leq \Gamma'(t)$. Integrating from $a$ to $x$ and using the FTC (the baby Riemann version) shows that \begin{align} \int_a^x|\gamma'(t)|\,dt&\leq \int_a^x\Gamma'(t)\,dt=\Gamma(x)-\Gamma(a)=\Gamma(x), \end{align} since $\Gamma(a)=0$.
This proves the theorem under the extra hypothesis of $\Gamma$ being differentiable with Riemann-integrable derivative. It turns out that this is unnecessary, but actually proving the theorem without these assumptions is much harder and I do not know how to do it without Lebesgue integrals and the notion of absolute continuity, and the corresponding Lebesgue-integral version of the Fundamental theorem of calculus.
For completeness, here's how the proof goes. We only have to assume $\gamma:[a,b]\to\Bbb{R}^n$ is absolutely continuous (for short, $\gamma$ is AC). Note also that Wikipedia only defines absolute continuity for functions with $\Bbb{R}$ as the target space, but the same definition can be made for $\Bbb{R}^n$-valued functions. From here, it is not too hard to show that $\Gamma$ is also AC. Hence (see theorem 7.20 in Rudin's RCA, which again holds for $\Bbb{R}^n$-valued functions by slight modifications), both $\gamma$ and $\Gamma$ are differentiable a.e on $[a,b]$ and the FTC holds. At any point $t\in (a,b)$ where $\gamma$ and $\Gamma$ are differentiable, we prove as above that $|\gamma'(t)|\leq \Gamma'(t)$. Since this is true for a.e $t\in [a,b]$, the Lebesgue-version of the FTC now implies \begin{align} \int_a^x|\gamma'(t)|\,dt&\leq \int_a^x\Gamma'(t)\,dt=\Gamma(x)-\Gamma(a)=\Gamma(x), \end{align} since $\Gamma(a)=0$.
So, the proof idea is actually very simple, but getting the technical details right under the weak assumptions is the difficult part.
To summarize, one has the following strong theorem:
Theorem 1 is a special case of theorem 2 (since the hypotheses of theorem 1 imply those of theorem 2; see Rudin's RCA theorem 7.20 for this, and also because if the Riemann integral exists, so does the Lebesgue integral, and the two are equal).
In view of theorem 1 (the proof of which we had to take a long detour), it follows that there is no curve $\gamma$ satisfying the three conditions you pose.