First let us not assume that $X$ has rational singularities, just that it is a normal projective surface and $C\subset X$ a curve on $X$. Let $f:Y\to X$ be a resolution and
$\widetilde C=f^{-1}_*C\subset Y$ the strict transform of $C$ on $Y$.
Consider the following commutative diagram:
$$
0 \to \mathscr I_C \to \mathscr O_X \to \mathscr O_C \to 0\quad\qquad\qquad\qquad\qquad\qquad\qquad
$$
$$\alpha\downarrow\qquad \beta\downarrow \qquad \gamma\downarrow \qquad\qquad\qquad\qquad\qquad\qquad\qquad$$
$$\qquad
0 \to f_*\mathscr I_{\widetilde C} \to f_*\mathscr O_Y \to f_*\mathscr O_C \to R^1f_*\mathscr I_{\widetilde C}\to R^1f_*\mathscr O_{Y}\to 0
$$
Since $X$ is normal, $\beta$ is an isomorphism and it is clear that $\alpha$ and $\gamma$ are injective. Then by the Snake Lemma $\alpha$ is also an isomorphism and we obtain that we have an exact sequence
$$
0\to \mathscr O_C \to f_*\mathscr O_C \to R^1f_*\mathscr I_{\widetilde C}\to R^1f_*\mathscr O_{Y}\to 0
$$
Obviously the two $R^1$ sheaves are supported on a zero-dimensional scheme $P$, so we have
$$
\chi(\mathscr O_C)=\chi(\mathscr O_{\widetilde C})-\mathrm{length}(R^1f_*\mathscr I_{\widetilde C}) + \mathrm{length}(R^1f_*\mathscr O_{Y})
$$
Now
1 If $(X,C)$ is log canonical, then it is a DB pair by Thm 1.4 of this paper and Prop 5.1 of this paper. Then $R^1f_*\mathscr I_{\widetilde C}=0$ by Cor 6.2 of the same paper. Therefore in this case $\chi(\mathscr O_C)=\chi(\mathscr O_{\widetilde C})$. But of course, I just realize now that this simply means that if $(X,C)$ is lc, then at every point either $X$ is smooth or $C$ is smooth, so this is not a big surprise.
2 If $X$ has rational singularities, then $R^1f_*\mathscr O_{Y}=0$, so the question reduces to determining $R^1f_*\mathscr I_{\widetilde C}$. In this case this is the same as Sasha's sheaf $T$. Clearly, it feels that one should be able to compute this by knowing the intersection of the exceptional set with $\widetilde C$.
In these situations one may try to use the Theorem on Formal Functions. That says that if $E$ denotes the pre-image of the reduced scheme supported on the singular set of $X$ (i.e., for each singular point the pre-image of the closed point defined by the maximal ideal, sometimes this is called the Artin cycle), then
$$
(R^1f_*\mathscr I_{\widetilde C})_P^{\wedge}\simeq \underset{\leftarrow}{\lim}\ H^1(mE,\mathscr I_{\widetilde C}\otimes \mathscr O_{mE})
$$
Fortunately the rest of the computation is happening on $Y$ so every divisor is Cartier.
I would try to compute the right hand side using the short exact sequences
$$
0\to \mathscr L^{m} \to \mathscr O_{(m+1)E} \to \mathscr O_{mE} \to 0 \tag{$\star$}
$$
where $\mathscr L=\mathscr I_{E}/\mathscr I_{E}^2=\mathscr N_{E/Y}^{-1}$ the dual of the normal bundle of $E$ which is indeed a line bundle since $E$ is a local complete intersection and for the same reason $\mathscr I_{E}^m/\mathscr I_{E}^{m+1}\simeq \mathscr L^m$.
Again, $\mathscr I_{\widetilde C}$ is a line bundle on $Y$ and remains that when restricted to $E$, so $(\star)$ induces the short exact sequence
$$
0\to \mathscr I_{\widetilde C}\otimes\mathscr L^{m} \to \mathscr I_{\widetilde C}\otimes\mathscr O_{(m+1)E} \to \mathscr I_{\widetilde C}\otimes\mathscr O_{mE} \to 0 $$
Now this is where you will need some specific information. The increment in the above inverse limit is given by the cokernel of the edge map
$$
H^0(E, \mathscr I_{\widetilde C}\otimes\mathscr O_{mE})\to
H^1(E, \mathscr I_{\widetilde C}\otimes\mathscr L^{m})
$$
$\mathscr L$ is ample on $E$, so by Serre vanishing $H^1(E, \mathscr I_{\widetilde C}\otimes\mathscr L^{m})=0$ for $m\gg 0$, so this is indeed a finite game.
Another thing that could help is that
by Thms 3 and 4 of this paper $h^0(E,\omega_E)=0$ and hence
$h^1(E, \mathscr I_{\widetilde C}\otimes\mathscr L^{m}) =h^0(E,\omega_E\otimes \mathscr I_{\widetilde C}^{-1}\otimes\mathscr L^{-m})=0$ as soon as $\mathscr L^m(-\widetilde C)$ has a global section. You should be able to figure out the first $m$ for which this happens as that depends on $E^2$ (see Thm 4 of ibid) and $\widetilde C\cdot E$. For the actual dimension of the intermediate steps you might be able to use some of Artin's methods from here.
This has certainly grown to a much longer answer than I anticipated when I started and although it does not give you a complete answer it might give you some ideas to go on.
