I'd prefer to work in homology (with coefficients $\mathbb{Q}$ everywhere). I'll say that a class $x\in H_k(X)$ is basically toral if there is a map $f\colon T^k\to X$ sending the fundamental class of $T^k$ to $x$. I'll say that $x$ is toral if it is a linear combination of basically toral classes, and I'll write $T_*(X)$ for the group of toral classes. This is clearly a subfunctor of $H_*(X)$, and $T^*(X)$ is the complement of the annihilator of $T_*(X)$. I'll say that $X$ is toral if $T_*(X)=H_*(X)$. It is easy to see that $T^n$ is toral and thus that $T_*(X)$ is the sum of the images $f_*(H_*(T^n))$ for all maps $f\colon T^n\to X$. This makes it clear that $T_*(X)$ is a subcoalgebra of $H_*(X)$, and also that $T_*(X\times Y)$ contains $T_*(X)\otimes T_*(Y)$. From this it follows that the class of toral spaces is closed under products. It is also easily seen to be closed under disjoint unions. Also, if $f\colon X\to Y$ and $X$ is toral and $f_*\colon H_*(X)\to H_*(Y)$ is surjective then $Y$ is toral. Using this we see that the class of toral spaces is closed under wedges and smash products.
Note also that if $G$ is a connected Lie group and $T$ is a maximal torus in a maximal compact subgroup of $G$, then the inclusion $i\colon T\to G$ gives a surjection of rational homology groups. It follows that all connected Lie groups are toral, and the connectivity constraint is easily removed.
Next, there is an $H_*$-epimorphism $T^n\to S^n$, so spheres are toral, so products of spheres are toral.
Now let $X$ be a connected toral space. Using the James construction, we see that there is a family of maps $X^k\to\Omega\Sigma X$ that are jointly surjective in homology, so $\Omega\Sigma X$ is again toral. There is a standard map $\Omega\Sigma S^2\to\mathbb{C}P^\infty$ that gives an isomorphism in rational homology, so the space $\mathbb{C}P^\infty=BU(1)$ is toral. It follows that for any $n$ the space $BT^n=BU(1)^n$ is toral. By considering maximal tori again, we see that $BG$ is toral for any connnected Lie group $G$. In particular, the infinite Grassmannian $BU(n)=\text{Grass}_n(\mathbb{C}^\infty)$ is toral.
Now suppose that $X$ is toral and that $Y$ is a homological $d$-skeleton of $X$, meaning that $Y$ is a $d$-skeleton with respect to some CW structure and that the map $i_*\colon H_*(Y)\to H_*(X)$ is injective. If $y\in H_k(Y)$ then wlog $k\leq d$ and we can express $i_*(y)$ as a linear combination of terms carried by maps $f_j\colon T^k\to X$. All these maps $f_j$ can be deformed so that they land in $Y$, and using this, we see that $y$ is toral. As $y$ was arbitrary we deduce that $Y$ is toral. In particular, we can take $X=\mathbb{C}P^\infty$ to see that the space $\mathbb{C}P^m=\text{Grass}_1(\mathbb{C}^{m+1})=U(m+1)/(U(1)\times U(m))$ is toral. I would guess that all homogeneous spaces $G/H$ (with $G$ a compact Lie group and $H$ a closed subgroup) are toral, but I do not immediately see a proof. This would include the finite Grassmannians $G_k(\mathbb{C}^m)$ for example. Toric varieties would be another interesting test case.
Now let $X$ be a connected infinite loop space, so $X\simeq\Omega^\infty T$ for some $0$-connected spectrum $T$. By standard arguments in stable homotopy, we can choose a wedge of spheres $W$ and a map $f\colon W\to T$ that gives an isomorphism $\pi_*(W)\otimes\mathbb{Q}\to\pi_*(T)\otimes\mathbb{Q}$, or equivalently an isomorphism $H_*(W)\to H_*(T)$. This gives a map from the space $Y=\Omega^\infty W$ to $X$ that is surjective in rational homology. Now $Y$ is a filtered colimit of finite products of spaces of the form $QS^{2n+1}$ or $QS^{2n+2}$. There are $H_*$-epimorphisms $S^{2n+1}\to QS^{2n+1}$ and $\Omega\Sigma S^{2n+2}\to QS^{2n+2}$, and using these we see that $Y$ is toral, and thus that $X$ is toral. Again, the connectivity constraint is easily removed.
For an interesting example in the opposite direction, I would like to consider the configuration space $F_n\mathbb{C}$ of $n$-tuples of distinct points in $\mathbb{C}$. There is a well-known calculation of the cohomology: it has generators $a_{pq}\in H^1$ for $1\leq p,q\leq n$ subject to $a_{pp}=0$ and $a_{pq}=a_{qp}$ and $a_{pq}^2=0$ and $a_{pq}a_{qr}+a_{qr}a_{rp}+a_{rp}a_{pq}=0$. The space can also be described as $B\Gamma_n$, where $\Gamma_n$ is the braid group on $n$ strings, so $[T^k,F_n\mathbb{C}]=\text{Hom}(\mathbb{Z}^k,\Gamma_n)/\text{conjugacy}$. From this one can probably work out $T_*(F_n\mathbb{C})$ and prove that it is not equal to $H_*(F_n\mathbb{C})$ when $n>2$, but I am not sure of the details. Other discrete groups such as mapping class groups, automorphisms of free groups and $GL_n(\mathbb{Z})$ may also be interesting test cases. The question mentions compact oriented surfaces, which can also be seen as classifying spaces of discrete groups.
Along the same lines, we can describe $T^n\vee T^n$ as $B(\mathbb{Z}^n*\mathbb{Z}^n)$ (where $*$ denotes free product of groups). As the free product is highly non-commutative, there are not so many homomorphisms $\mathbb{Z}^m\to\mathbb{Z}^n*\mathbb{Z}^n$. Using this, I think we can show that the sum of the two top classes in $H_n(T^n\vee T^n)$ is not basically toral, so we do not get a good theory by restricting to the basically toral case.
