Take $s \ge 1$.   We will use $(x+y)^s \le x^s+y^s$ for positive $x,y$.

Let $C=2^{1+s/2}(1+c)$. I claim: $\mathcal H^s(G) \le C$.    

Let $N \in \mathbb N$.  Let $\eta > 0$ be so small that
$N2^s\eta^s \le c$.

For $j=1,2,\dots$ let
$$
M_j = \sup\left\{f(x): \frac{j-1}{N} \le x \le \frac{j}{N}\right\},\qquad
m_j = \inf\left\{f(x): \frac{j-1}{N} \le x \le \frac{j}{N}\right\} .
$$


Now construct a partition $\mathcal P$ of $[0,1]$.  First put $0,1 \in \cal P$.  Then, for each $j$ from $1$ to $N$, put $x_j, y_j \in \cal P$, 
chosen so that $|f(x_j) - M_j| < \eta$ and
$|f(y_j) - m_j| < \eta$.  Then 
$|M_j - m_j| \le |f(x_j)-f(y_j)| + 2 \eta$.  This partition has $2N+2$ dividing points in it.  Estimate
\begin{align}
\sum_{j=1}^N |M_j-m_j|^s &\le \sum_{j=1}^N\big(|f(x_j)-f(y_j)|+2 \eta\big)^s
\\ &\le N2^s\eta^s+\sum_{j=1}^N |f(x_j))-f(y_j)|^s \le
N2^s\eta^s+c \le 2c .
\end{align}

Now construct a cover $\mathcal Q$ of $G$.  Subdivide the plane into squares of side $1/N$ using vertical lines $x=k/N, k \in \mathbb Z$ and horizontal lines $y=k/N, k \in \mathbb Z$.  Let $\mathcal Q$ include those squares that meet
the graph $G$.  Let us say an index $j$ is **normal** if $M_j - m_j < 1/N$, and **abnormal** if $M_j - m_m \ge 1/N$

For a given $j$, how many squares in $\mathcal Q$ are there with base $[(j-1)/N,j/N]$?  If $j$ is normal, then there are at most $2$.  If $j$ is abnormal
then there are at most
$$
\lceil NM_j\rceil - \lfloor Nm_j\rfloor \le N(M_j-m_j) +2 .
$$
Now compute
$$
\sum_{U \in \mathcal Q} \text{diam}(U)^s \le
\left(\frac{\sqrt2}{N}\right)^s \left[\sum_{j\text{ normal}} 2+
\sum_{j\text{ abnormal}}\big( N(M_j-m_j) +2\big)\right]
$$
But
$\sum_{j=1}^N 2 = 2N$ and
$$
\sum_{j\text{ abnormal}} N(M_j-m_j) 
\le N \sum_{j\text{ abnormal}} (M_j-m_j)^s N^{s-1}
\le N^s \sum_{j=1}^N (M_j - m_j)^s \le 2cN^s .
$$
Therefore
$$
\sum_{U \in \mathcal Q} \text{diam}(U)^s \le 
\frac{2^{s/2}}{N^s}\big[2N + 2cN^s\big]
=\frac{2^{1+s/2}}{N^{s-1}} + 2^{1+s/2} c \le 2^{1+s/2} (1+c) = C
$$

Hence $\mathcal H^s_{\sqrt{2}/N} \le C$.  

This is true for all (large enough) $N$, so $\mathcal H^s(G) \le C$.