twisted Poisson structures, degenerate metrics and integrability properties of (2,0)-tensors Given a regular (constant rank) bi-vector $\Pi \in \Gamma(\bigwedge^2TM)$ on a smooth manifold $M$ the necessary and sufficient condition for the image of $\Pi^\sharp:T^*M\to TM$ to be an integrable distribution is that $\Pi$ is a twisted Poisson tensor with respect to a closed 3-form $\phi$, i.e. there exist a closed 3-form $\phi$ such that
$$[\Pi,\Pi]=\Pi^\sharp(\phi)$$
(see e.g. arXiv:1104.0880).
What is the corresponding condition for a degenerate regular symmetric (2,0)-tensor (a degenerate metric) $g\in \Gamma(S^2TM)$ to produce an integrable distribution as the image of $g^\sharp:T^*M\to TM$?
And what about any regular (constant rank) tensor $T\in \Gamma(TM\otimes TM)$, with no special symmetry properties?
 A: Perhaps you will find this a useful answer, at least for the symmetric case:  Let $S^2_r(TM)\subset S^2(TM)$ denote the subbundle consisting of the cometrics of rank $r$.  (This bundle is the disjoint union of $r{+}1$ smooth subbundles that are distinguished by the algebraic type of the symmetric bivector.  However, this 'type division' won't play a role in this analysis.)  
There is a natural first-order differential operator
$$
\delta_r:C^\infty\bigl(S^2_r(TM)\bigr)\longrightarrow 
C^\infty\bigl(\Lambda^2(TM)\otimes \Lambda^r(TM)\otimes \Lambda^{r+1}(TM)\bigr)
$$
with the property that $\delta_r(g)=0$ if and only if the $r$-rank subbundle $g^\sharp(T^\ast M)\subset  TM$ is integrable.  
This differential operator is defined as follows: Locally, $g$ can be written in the form
$$
g = \sum_{i,j=1}^rg^{ij}\ X_iX_j\ ,
$$
where $X_1,\ldots, X_r$ are linearly independent vector fields, $g^{ij}=g^{ji}$ are functions, and $\Delta = \det (g^{ij})$ is nonvanishing.  Now define (using the summation convention)
$$
\delta_r(g) = \Delta\ g^{ik}g^{jl}\ \left(X_i{\wedge}X_j\ \otimes\ X_1{\wedge}X_2{\wedge}\cdots{\wedge}X_r 
\ \otimes\ [X_k,X_l]{\wedge}X_1{\wedge}X_2{\wedge}\cdots{\wedge}X_r\right).
$$
It is easy to verify that this is independent of the choice of basis $X_i$, and so it is globally defined.  It is also obvious that it vanishes if and only if the bundle $g^\sharp(T^\ast M)$ (which is spanned locally by the $X_i$) is integrable.
Note that $\delta_r$ vanishes identically unless $2 \leq r < n$, as would be expected.  Also, $\delta_r(\lambda g) = \lambda^{r+2}\ \delta_r(g)$ for all nonvanishing functions $\lambda$.
If I had considered oriented cometrics of rank $r$ (i.e., I had, in addition, fixed an orientation of $g^\sharp(T^\ast M)$), then I could then have got rid of the $\Lambda^r(TM)$ factor in the definition of $\delta_r$.  However, I didn't want to impose the orientability condition, and the $\Lambda^r(TM)$ factor fixes that problem. 
The almost Poisson case:  Although you already have a criterion in the almost Poisson case, note that there is a similar construction to test the integrability of the bundle $\Pi^\sharp(T^\ast M)$ in the case of an almost Poisson structure $\Pi$ of constant half-rank $r$:  Let $\Lambda^2_r(TM)\subset \Lambda^2(TM)$ be the subbundle of bivectors of half-rank $r$, i.e., its sections are those $\Pi$ such that $\Pi^r$ is nonvanishing as a section of $\Lambda^{2r}(TM)$ while $\Pi^{r+1}\equiv0$.  
There is a natural first-order differential operator 
$$
\partial_r:C^\infty\bigl(\Lambda^2_r(TM)\bigr)\longrightarrow 
   C^\infty\bigl(\Lambda^2(TM)\otimes\Lambda^{2r+1}(TM)\bigr)
$$
defined as follows:  Locally, write $\Pi$ in the form
$$
\Pi = \sum_{i,j=1}^{2r} {\tfrac12}\  a^{ij}\ X_i\wedge X_j\ ,
$$
where $X_1,\ldots,X_{2r}$ are linearly independent vector fields, $a^{ij}=-a^{ji}$ are functions, and $\mathrm{Pf}(a^{ij})$ is nonvanishing (since $\Pi^r = r!\ \mathrm{Pf}(a^{ij})\ X_1\wedge X_2\wedge\cdots\wedge X_{2r}\not=0$).  Now define (again using the summation convention)
$$
\partial_r(\Pi) = 
  {\tfrac14}\ a^{ik}a^{jl}\ 
   X_k{\wedge}X_l\otimes\ [X_i, X_j]{\wedge}\Pi^r\ .
$$
It is easy to verify that this does not depend on the choice of basis $X_i$, 
and so it is well-defined.  
It is evident that that $\partial_r(\Pi)$ vanishes if and only if the bundle $\Pi^\sharp(T^\ast M)$ (which is spanned locally by the $X_i$) is integrable. Also, $\partial_r(\lambda\ \Pi)= \lambda^{r+2}\partial_r(\Pi)$ for all nonvanishing functions $\lambda$.
This is a somewhat more explicit test that the one you proposed, I think.
The general case:  It turns out that one does not need any assumption of symmetry or skew-symmetry to get the integrability tensor, just an assumption of constant rank.  Let $\otimes^2_r(TM)\subset TM\otimes TM$ denote the subbundle consisting of the tensors of rank $r$.  A section $\phi\in C^\infty\bigl(\otimes^2_r(TM)\bigr)$ can be written locally in the form 
$$
\phi = \sum_{i,j=1}^r f^{ij}\ X_i\otimes Y_j
$$
where $X_1,\ldots,X_r$ are linearly independent vector fields, $Y_1,\ldots,Y_r$ are linearly independent vector fields, and $f=(f^{ij})$ is an invertible matrix, i.e., $\det(f)\not=0$.  The vector fields $X_i$ are local sections of a globally defined bundle $\lambda_\phi\subset TM$ of rank $r$, and the vector fields $Y_i$ are local sections of a globally defined bundle $\rho_\phi\subset TM$ of rank $r$.
There is a natural first-order differential operator
$$
D_r: C^\infty\bigl(\otimes^2_r(TM)\bigr)\longrightarrow
     C^\infty\bigl(\Lambda^2(TM)\otimes \Lambda^r(TM)\otimes \Lambda^{r+1}(TM)\bigr)
$$
which, relative to the local expression given above, takes the form
$$
D_r(\phi) = 
\det(f)\ f^{ik}f^{jl}\ \left(Y_k{\wedge}Y_l\ \otimes
\ Y_1{\wedge}Y_2{\wedge}\cdots{\wedge}Y_r 
\ \otimes\ [X_i,X_j]{\wedge}X_1{\wedge}X_2{\wedge}\cdots{\wedge}X_r\right).
$$ 
This operator has the property that $D_r(\phi)=0$ if and only if $\lambda_\phi$ is integrable.  
Note that there is a well-defined involution $\iota$ on $C^\infty\bigl(\otimes^2_r(TM)\bigr)$ such that
$$
\iota(\phi) = \sum_{i,j=1}^r f^{ij}\ Y_j\otimes X_i\ ,
$$
and the integrability of $\rho_\phi$ is equivalent to the equation $D_r\bigl(\iota(\phi)\bigr)=0$.
A: This is also only for the symmetric case.
Let $S(TM)=\bigoplus_k S^k(TM)$ and $\Gamma(S(TM))$ be the commutative graded algebra of symmetric purely contravariant tensor fields. This is the algebra of functions on $T^*M$ which are polynomial on each fiber, and it is invariant under the Poisson bracket on $T^*M$.
This is the symmetric Schouten bracket.
See the following paper for more information: 
Michel Dubois-Violette, Peter W. Michor: A common generalization of the Frölicher-Nijenhuis bracket and the Schouten bracket for symmetric multi vector fields, Indagationes Math. N. S. 6 (1995), 51--66.
(pdf)
It might be that one can just carry over the considerations from $\Gamma(\Lambda (TM))$
to $\Gamma(S(TM))$ using this bracket.
EDIT: The above did not work. So let me try again. I do not need symmetry now.
Suppose that $a\in\Gamma(TM\otimes TM)$ is of constant rank when viewed as 
$a:T^\star M\to TM$.
Choose a Riemann metric $g$ on $M$ and consider $A=a\circ g:TM\to T^\star M\to TM$ which is again of constant rank. So $\ker(A)$ and $A(TM)$ are subbundles. Choose an isomorphism $B:TM\to TM$
such that $A.B$ equals the $g$-orthogonal projection $P$ onto $A(TM)$. $P$ can be canonically constructed from $g$ and $a$, up to some free choice of an isomorphism between $Im(A)^\bot$ and $\ker(A)$. 

Now the Frölicher-Nijenhuis bracket of the vector valued 1-form $P$ is of the form
$$ [P,P] = 2R + 2\bar R,\quad R(X,Y)= P[(Id-P)X,(Id-P)Y], \quad \bar R(X,Y)=(Id-P)[PX,PY]
$$
$R$, the curvature, is exacly the obstruction against integrability of the kernel of $P$.

$\bar R$, the cocurvature, is exactly the obstruction against integrability of the image of $P$, which we are interested in. 

So the condition is $\bar R=0$ which can be written as 
$$
0=P^*[P,P]=[P,P].(P\times P):\Lambda^2TM \to TM.
$$
A: i might have a partial answer, at least for the symmetric case. however my reasoning might not be completely stringent, so i still would like to read a better argument.
Let $X,Y\in \Gamma(Image(g^{\sharp}))$, then $X^i=g^{ik}\alpha_k$, $Y^i=g^{ik}\beta_k$ with $\alpha, \beta$ being any two 1-forms. Let us use Frobenius theorem to study the integrability of $Image(g^\sharp)\subset TM$ and compute:
$$[X,Y]^j=X^k \frac{\partial Y^j}{\partial x^k}-Y^k \frac{\partial X^j}{\partial x^k} = $$
$$=g^{jl} g^{ks} (\alpha_s \frac{\partial \beta_l}{\partial x^k}-\beta_s \frac{\partial \alpha_l}{\partial x^k})+\alpha_p \beta_l (g^{pk}\frac{\partial g^{jl}}{\partial x^k}-g^{lk}\frac{\partial g^{jp}}{\partial x^k})=$$
$$=g^{jl} g^{ks} (\alpha_s \frac{\partial \beta_l}{\partial x^k}-\beta_s \frac{\partial \alpha_l}{\partial x^k})+\alpha_p \beta_l (\Gamma^{plj}-g^{jk}\frac{\partial g^{pl}}{\partial x^k})$$
Now this last expression is in the image of $g^\sharp$ if and only if the (completely contravariant) Christoffel symbol $\Gamma^{plj}$ factors as $\tilde{\Gamma}^{pl}_r g^{rj}$. Now, here I'm not quite certain, but doesn't this mean that the (contravariant) Levi-Civita connection of $g$ in $TM$ must reduce to a connection on the vector bundle $Image(g^\sharp)\subset TM$?
