L. Schwartz and A. Grothendieck made clear, by very early 1950s, that the Cauchy (-Goursat) theory of holomorphic functions of a single complex variable extended with essentially no change to functions with values in a locally convex, quasi-complete topological vector space. Cauchy integral formulas, residues, Laurent expansions, etc., all succeed (with trivial modifications occasionally).
Conceivably one needs a little care about the notion of "integral". The Gelfand-Pettis "weak" integral suffices, but/and a Bochner version of "strong" integral is also available.
Further, in great generality, as Grothendieck made clear, "weak holomorphy" (that is, $\lambda\circ f$ holomorphic for all (continuous) linear functionals $\lambda$ on the TVS) implies ("strong") holomorphy (i.e., of the TVS-valued $f$).
(Several aspects of this, and supporting matter, are on-line at http://www.math.umn.edu/~garrett/m/fun/Notes/09_vv_holo.pdf and other notes nearby on http://www.math.umn.edu/~garrett/m/fun/)
Edit: in response to @Christopher A. Wong's further question... I've not made much of a survey of recent texts to see whether holomorphic TVS-valued functions are much discussed, but I would suspect that the main mention occurs in the setting of resolvents of operators on Hilbert and Banach spaces, abstracted just a little in abstract discussions of $C^*$ algebras. (Rudin's "Functional Analysis" mentions weak integrals and weak/strong holomorphy and then doesn't use them much, for example.) Schwartz' original book did treat such things, and was the implied context for the first volume of the Gelfand-Graev-etal "Generalized Functions". In the latter, the examples are very small and tangible, but (to my taste) tremendously illuminating about families of distributions.
Edit-edit: @barto's further question is about the behavior of holomorphic operator-valued $f(z)$ at an isolated point $z_o$ where $f(z)$ fails to be invertible. I do not claim to have a definitive answer to this, but only to suggest that the answer may be complicated, since already for the case $f(z)=(T-z)^{-1}$ for bounded, self-adjoint $T$, it seems to take a bit of work (the spectral theorem) to prove that isolated singularities are in the discrete/point spectrum of $T$. But this may be overkill, anyway...
Let $F : \mathcal{O}\subset\mathbb{C}\rightarrow \mathcal{L}(X,Y)$ where $\mathcal{O}$ is an open subset of $\mathbb{C}$, and $\mathcal{L}(X,Y)$ denotes the bounded linear operators from a complex Banach space $X$ to another complex Banach space $Y$. Then $F$ is holomorphic iff the following limits exist for all $z\in\mathcal{O}$:
$$
\lim_{w\rightarrow z}\frac{1}{w-z}(F(w)-F(z)).
$$
In that case, the limit is denoted by $F'(z)$.
It can be proved that $F$ is holomorphic iff $z\mapsto F(z)x$ is a holomorphic vector function for all $x\in X$. And $F(z)x$ is holomorphic iff $z\mapsto y^*(F(z)x)$ is a complex holomorphic function for all $y^*\in Y^*$. The proof follows from the uniform boundedness principle, and a power series argument. This remarkable bootstrapping is due to the Uniform Boundedness Principle and a power series argument, where the existence of a complex derivative of a complex function $f : \mathcal{O}\rightarrow\mathbb{C}$ amounts to boundedness of the following for all $w$ in a punctured neighborhood of $z$:
$$
\frac{1}{w-z}\left[\frac{f(w)-f(z)}{w-z}-f'(z)\right],\;\;\;w\in B_r(z)\setminus \{z\}.
$$
Best Answer
The map $T$ here is really not relevant; you might as well just replace $\Gamma$ with $T\circ\Gamma$ at the start. That is, suppose $\Gamma:\mathbb{D}\to Y$ is holomorphic. Then you want to find a holomorphic $f:\mathbb{D}\to Y$ such that $f(0)=0$ and $f'=\Gamma$. You can do this in the same way as in the case $Y=\mathbb{C}$: just integrate. That is, define $$f(z)=\int_0^z\Gamma(s)\,ds$$ where the integral is a Bochner integral of $Y$-valued functions along some path from $0$ to $z$. To prove that $f'=\Gamma$, you can first show that the integral above is independent of the path chosen by composing with an arbitrary bounded linear functional $\alpha:Y\to\mathbb{C}$ to reduce to the case $Y=\mathbb{C}$ (as discussed in this answer). Once you have that, you can prove that $f'=\Gamma$ by exactly the same argument as in the case $Y=\mathbb{C}$.