[Added: I misread the question, and in fact this answer does not answer the OP's question, but rather the following question: is $\phi(T)$ closed in Spec $\Gamma(T)$, which is a different question. Probably the upvotes can be attributed to the link to the stacks project!]
If $T$ is quasi-affine (i.e. admits an open immersion into affine space), then the map $\phi$ is an open immersion, and in fact Spec $\Gamma(T)$ is the initial object in the category of affine schemes containing $T$ as an open subscheme.
In particular, in this case $\phi$ has closed image if and only if and only if $T$ is in fact affine. [Added: As Qing Liu points out in a comment below, in this quasi-affine situation, $\phi$ is in fact a closed map onto its image.]
Thus if we take $T$ to be ${\mathbb A}^2_k \setminus \{0\}$ for some field $k$, i.e. affine $2$-space with the origin removed, then we get an example of $T$ where this map is open with non-closed image (since this $T$ is quasi-affine but not affine). Note that Spec $\Gamma(T) = {\mathbb A^2}_k$.
(This is a geometric analogue of Qing Lui's more arithmetic example; what both have in common is that a closed point was removed from a 2-dimensional affine scheme, so as to make a quasi-affine scheme that is not affine.[Added: I also misread Qing Liu's example; my remark would apply to the affine line over ${\mathbb Z}$ with a closed point removed; Qing's example is more complicated, since it is actually dealing with the OP's question.]question. One can make a geometric analogue of Qing's example by deleting a closed point from ${\mathbb A}^1\times {\mathbb P}^1$; more geometrically still, remove one of the lines of a ruling from a projective quadric surface, and then remove an additional point.])
EDIT: In the definition of quasi-affine, one should also require that $T$ be quasi-compact. (The stacks project is a terrific resource for these foundational definitions in scheme theory, particularly with regard to finiteness and separation issues.)
Note that if $T$ is any quasi-compact scheme, then the map $T \to$ Spec $\Gamma(T)$ has dense image. (If $f \in \Gamma(T)$ and $D(f)$ is the usual affine open in Spec $\Gamma(T)$, i.e. Spec $\Gamma(T)_f$, then if $\phi^{-1}(D(f))$ is empty, it must be that $f$ is locally nilpotent on $T$. Since $T$ is quasi-compact this implies that $f$ is actually nilpotent, and hence that $D(f)$ is empty.) As Martin notes in his answer, this is similarly true if $T$ is reduced.
It need not be true if $T$ is non-reduced and non-quasi-compact (since $T$ may then admit locally nilpotent sections of $\mathcal O_T$ that are not globally nilpotent, e.g. $T = \coprod_n$ Spec $k[x]/(x^n)$).
